Subjective Questions
Q1:
Prove that:
\(\frac {sin\theta + sin2\theta}{1 + cos\theta + cos2\theta}\) = tan\(\theta\)
Type: Short
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=\(\frac {sin\theta + sin2\theta}{1 + cos\theta + cos2\theta}\)</p> <p>= \(\frac {sin\theta + 2sin\theta cos\theta}{1 + cos\theta + 2cos^2\theta - 1}\)</p> <p>= \(\frac {sin\theta (1 + 2cos\theta)}{cos\theta (1 + 2cos\theta)}\)</p> <p>= \(\frac {sin\theta}{cos\theta}\)</p> <p>= tan\(\theta\)</p> <p>Hence, L.H.S. = R.H.S. <sub>proved</sub></p>
Q2:
Express sin3A cos2A in terms of sin A.
Type: Short
Difficulty: Easy
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Answer: <p>Here,</p> <p>sin3A cos<sup>2</sup>A</p> <p>= (3 sinA - 4 sin<sup>3</sup>A) (1 - sin<sup>2</sup>A)</p> <p>= 3 sinA - 4 sin<sup>3</sup>A - 3 sin<sup>3</sup>A + 4 sin<sup>5</sup>A</p> <p>= 4 sin<sup>5</sup>A - 7 sin<sup>3</sup>A + 3 sinA <sub>Ans</sub></p>
Q3:
Prove that:
cos 2A = \(\frac {1 - tan^2A}{1 + tan^2A}\)
Type: Short
Difficulty: Easy
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Answer: <p>Here,</p> <p>L.H.S.</p> <p>= cos 2A</p> <p>= \(\frac {cos^2A - sin^2A}{1}\)</p> <p>= \(\frac {cos^2A - sin^2A}{cos^2A + sin^2A}\)</p> <p>=\(\frac {{\frac {cos^2A}{sin^2A}} - {\frac {sin^2A}{cos^2A}}}{{\frac {cos^2A}{cos^2A}} + {\frac {sin^2A}{cos^2A}}}\)</p> <p>= \(\frac {1 - tan^2A}{1 + tan^2A}\)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q4:
If tan A = \(\frac 34\), find the value of sin 2A and cos 2A.
Type: Short
Difficulty: Easy
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Answer: <p>Here,</p> <p>tan A = \(\frac 34\)</p> <p>sin 2A</p> <p>= \(\frac {2tan A}{1 + tan^2A}\)</p> <p>= \(\frac {2 × \frac 34}{1 + (\frac {3}{4})^2}\)</p> <p>= \(\frac {\frac 32}{1 + \frac {9}{16}}\)</p> <p>= \(\frac 32\)× \(\frac {16}{25}\)</p> <p>= \(\frac {24}{25}\) <sub>Ans</sub></p> <p>cos 2A</p> <p>= \(\frac {1 - tan^2A}{1 + tan^2A}\)</p> <p>= \(\frac{1 -(\frac {3}{4})^2}{1 + (\frac {3}{4})^2}\)</p> <p>= \(\frac {1 - \frac 9{16}}{1 +\frac 9{16}}\)</p> <p>= \(\frac {\frac {16 - 9}{16}}{\frac {16 + 9}{16}}\)</p> <p>= \(\frac {7}{16}\)× \(\frac {16}{25}\)</p> <p>= \(\frac 7{25}\) <sub>Ans</sub></p>
Q5:
Prove that:
\(\frac {sin 2A}{1 + cos 2A}\) = tan A
Type: Short
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=\(\frac {sin 2A}{1 + cos 2A}\)</p> <p>= \(\frac {2 sin A cos A}{2 cos^2A}\)</p> <p>= \(\frac {sin A}{cos A}\)</p> <p>= tan A</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q6:
If cos\(\alpha\) = \(\frac {\sqrt 3}{2}\), find the value of sin3\(\alpha\) and cos3\(\alpha\).
Type: Short
Difficulty: Easy
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Answer: <p>Here,</p> <p>cos\(\alpha\) = \(\frac {\sqrt 3}{2}\)</p> <p>sin\(\alpha\)</p> <p>= \(\sqrt {1 - cos^2\alpha}\)</p> <p>= \(\sqrt {1 - (\frac {\sqrt 3}{2})^2}\)</p> <p>=\(\sqrt {1 - \frac 34}\)</p> <p>=\(\sqrt {\frac {4 - 3}{4}}\)</p> <p>=\(\sqrt {\frac 14}\)</p> <p>= \(\frac 12\)</p> <p>sin3\(\alpha\)</p> <p>= 3 sin\(\alpha\) - 4 sin<sup>3</sup>\(\alpha\)</p> <p>= 3× \(\frac 12\) - 4× (\(\frac 12\))<sup>3</sup></p> <p>= \(\frac 32\) - \(\frac 48\)</p> <p>= \(\frac 32\) - \(\frac 12\)</p> <p>= \(\frac {3 - 1}{2}\)</p> <p>= \(\frac 22\)</p> <p>= 1 <sub>Ans</sub></p> <p>Again,</p> <p>cos 3\(\alpha\)</p> <p>= 4 cos<sup>2</sup>\(\alpha\) - 3 cos\(\alpha\)</p> <p>= 4 (\(\frac {\sqrt 3}{2}\))<sup>3</sup>- 3× \(\frac {\sqrt 3}{2}\)</p> <p>= \(\frac {12\sqrt 3}{8}\) - \(\frac {3\sqrt 3}{2}\)</p> <p>= \(\frac {3\sqrt 3}{2}\) - \(\frac {3\sqrt 3}{2}\)</p> <p>= 0 <sub>Ans</sub></p> <p></p>
Q7:
Prove that:
4 (cos310° + sin320°) = 3 (cos 10° + sin 20°)
Type: Short
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=4 (cos<sup>3</sup>10° + sin<sup>3</sup>20°)</p> <p>=4cos<sup>3</sup>10° + 4sin<sup>3</sup>20°</p> <p>= [cos 3(10°) + 3 cos 10°] + [3 sin20° - sin 3(20°)]</p> <p>= cos 30° + 3 cos 10° + 3 sin 20° - sin 60°</p> <p>= \(\frac {\sqrt 3}{2}\) + 3 (cos 10° + sin 20°) - \(\frac {\sqrt 3}{2}\)</p> <p>=3 (cos 10° + sin 20°)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q8:
If cos\(\theta\) = \(\frac {12}{13}\) and sin\(\alpha\) = \(\frac 45\), find the values of sin 2\(\theta\) and cos 2\(\alpha\).
Type: Short
Difficulty: Easy
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Answer: <p>Here,</p> <p>cos\(\theta\) = \(\frac {12}{13}\) and sin\(\alpha\) = \(\frac 45\)</p> <p>sin\(\theta\)</p> <p>= \(\sqrt {1 - cos^2\theta}\)</p> <p>= \(\sqrt {1 - (\frac {12}{13})^2}\)</p> <p>= \(\sqrt {\frac {169 - 144}{169}}\)</p> <p>= \(\sqrt {\frac {25}{169}}\)</p> <p>= \(\frac 5{13}\)</p> <p>cos\(\alpha\)</p> <p>= \(\sqrt {1 - sin^2\alpha}\)</p> <p>= \(\sqrt {1 - (\frac {4}{5})^2}\)</p> <p>= \(\sqrt {\frac {25 - 16}{25}}\)</p> <p>= \(\sqrt {\frac 9{25}}\)</p> <p>= \(\frac 35\)</p> <p>∴ sin 2\(\theta\) = 2 sin\(\theta\) cos\(\theta\) = 2× \(\frac 5{13}\)× \(\frac {12}{13}\) = \(\frac {120}{169}\) <sub>Ans</sub></p> <p>∴ cos 2\(\alpha\) = 2 cos<sup>2</sup>\(\alpha\) - 1 = 2× \(\frac 35\)× \(\frac 35\) - 1 = \(\frac {18 - 25}{25}\) = -\(\frac 7{25}\) Ans</p> <p></p>
Q9:
Prove that:
tan 4\(\theta\) - tan 3\(\theta\) - tan\(\theta\) = tan 4\(\theta\) . tan 3\(\theta\) . tan\(\theta\)
Type: Short
Difficulty: Easy
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Answer: <p>Here,</p> <p>3\(\theta\) + \(\theta\) = 4\(\theta\)</p> <p>Putting tan on both,</p> <p>tan (3\(\theta\) + \(\theta\)) = tan 4\(\theta\)</p> <p>or, \(\frac {tan 3\theta + tan\theta}{1 - tan 3\theta tan\theta}\) = tan 4\(\theta\)</p> <p>or, tan 3\(\theta\) + tan\(\theta\) = tan 4\(\theta\) - tan\(\theta\) tan 3\(\theta\) tan 4\(\theta\)</p> <p>or,tan\(\theta\) tan 3\(\theta\) tan 4\(\theta\) =tan 4\(\theta\) -tan 3\(\theta\) -tan\(\theta\)</p> <p>Hence, L.H.S. = R.H.S. <sub>proved</sub></p>
Q10:
Prove that:
\(\frac {sin 5\theta}{sin \theta}\) - \(\frac {cos 5\theta}{cos \theta}\) = 4 cos 2\(\theta\)
Type: Short
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=\(\frac {sin 5\theta}{sin \theta}\) - \(\frac {cos 5\theta}{cos \theta}\)</p> <p>= \(\frac {sin 5\theta cos \theta - cos 5\theta sin \theta}{sin \theta cos \theta}\)</p> <p>= \(\frac {sin (5\theta - \theta)}{sin\theta cos\theta}\)</p> <p>= \(\frac {sin 4\theta}{sin\theta cos\theta}\)</p> <p>= \(\frac {sin 2(2\theta)}{sin\theta cos\theta}\)</p> <p>= \(\frac {2 sin 2\theta cos 2\theta}{sin\theta cos\theta}\)</p> <p>= \(\frac {2 × 2 sin\theta cos\theta cos2\theta}{sin\theta cos\theta}\)</p> <p>= 4 cos 2\(\theta\)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q11:
Prove that:
\(\frac {cos^3A + sin^3A}{cosA + sinA}\) = 1 - \(\frac 12\)sin 2A
Type: Short
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>= \(\frac {cos^3A + sin^3A}{cosA + sinA}\)</p> <p>= \(\frac {(cosA + sinA)(cos^2A + sin^2A - sinA cosA)}{cosA + sinA}\)</p> <p>= 1 - \(\frac {2 sinA cosA}{2}\)</p> <p>= 1 - \(\frac 12\)sin 2A</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q12:
Prove that:
\(\frac 1{tan3A - tanA}\) - \(\frac 1{cot3A - cotA}\) = cot2A
Type: Short
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=\(\frac 1{tan3A - tanA}\) - \(\frac 1{cot3A - cotA}\)</p> <p>=\(\frac 1{tan3A - tanA}\) - \(\frac 1{{\frac 1{tan3A}} - {\frac 1{tanA}}}\)</p> <p>=\(\frac 1{tan3A - tanA}\) - \(\frac {\frac 1{tanA - tan3A}}{tan3A tanA}\)</p> <p>=\(\frac 1{tan3A - tanA}\) - \(\frac {tan 3A tanA}{tanA - tan3A}\)</p> <p>= \(\frac 1{tan3A - tanA}\) +\(\frac {tan 3A tanA}{tan3A - tanA}\)</p> <p>= \(\frac {1 + tan 3A tanA}{tan3A - tanA}\)</p> <p>= \(\frac {\frac 1{tan3A - tanA}}{1 + tan3A tanA}\)</p> <p>= \(\frac 1{cot (3A - A)}\)</p> <p>= \(\frac 1{cot2A}\)</p> <p>= tan 2A</p> <p>Hence, L.H.S. = R.H.S. <sub>proved</sub></p>
Q13:
Prove that:
\(\frac {cos\theta}{cos\theta - sin\theta}\) - \(\frac {cos\theta}{cos\theta + sin\theta}\) = tan 2\(\theta\)
Type: Short
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=\(\frac {cos\theta}{cos\theta - sin\theta}\) -\(\frac {cos\theta}{cos\theta +sin\theta}\)</p> <p>= \(\frac {cos\theta (cos\theta + sin\theta) - cos\theta (cos\theta - sin\theta)}{(cos\theta - sin\theta) (cos\theta + sin\theta)}\)</p> <p>= \(\frac {cos^2\theta + cos\theta sin\theta - cos^2\theta + cos\theta sin\theta}{cos^2\theta - sin^2\theta}\)</p> <p>= \(\frac {2 sin\theta cos\theta}{cos^2\theta - sin^2\theta}\)</p> <p>= \(\frac {sin 2\theta}{cos 2\theta}\)</p> <p>= tan 2\(\theta\)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p> <p></p>
Q14:
Prove that:
\(\frac {1 - cos 2A + sin2A}{1 + cos 2A + sin2A}\) = tanA
Type: Short
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=\(\frac {1 - cos 2A + sin2A}{1 + cos 2A + sin2A}\)</p> <p>= \(\frac {2sin^2A + 2sinAcosA}{2cos^2A + 2sinAcosA}\) [\(\because\) 1 - cos2A = 2sin<sup>2</sup>A and 1 + cos2A = 2cos<sup>2</sup>A]</p> <p>= \(\frac {2sin A (sinA + cosA)}{2cosA (cosA + sinA)}\)</p> <p>= tanA</p> <p>Hence, L.H.S. = R.H.S. <sub>proved</sub></p>
Q15:
Prove that:
\(\frac {1 - tan^2(\frac {π}{4} - \theta)}{1 + tan^2(\frac {π}{4} - \theta)}\) = sin 2\(\theta\)
Type: Short
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=\(\frac {1 - tan^2(\frac {π}{4} - \theta)}{1 + tan^2(\frac {π}{4} - \theta)}\)</p> <p>= cos 2(\(\frac {π}{4} - \theta)\)</p> <p>= cos(\(\frac {π}{2} - 2\theta)\)</p> <p>= sin 2\(\theta\)</p> <p>hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q16:
Prove that:
\(\frac {sin 3A}{sinA}\) - \(\frac {cos 3A}{cosA}\) = 2
Type: Short
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=\(\frac {sin 3A}{sinA}\) - \(\frac {cos 3A}{cosA}\)</p> <p>= \(\frac {3 sinA - 4 sin^3A}{sinA}\) - \(\frac {4 cos^3A - 3 cosA}{cosA}\)</p> <p>= \(\frac {sinA (3 - 4 sin^2A)}{sinA}\) - \(\frac {cosA (4 cos^2A - 3)}{cosA}\)</p> <p>= 3 - 4 (1 - cos<sup>2</sup>A) - 4 cos<sup>2</sup>A + 3</p> <p>= 3 - 4 + 4 cos<sup>2</sup>A - 4 cos<sup>2</sup>A + 3</p> <p>= 6 - 4</p> <p>= 2</p> <p>Hence, L.H.S = R.H.S. <sub>Proved</sub></p>
Q17:
Prove that:
4 cosec 2A . cot 2A = cosec2A - sec2A
Type: Short
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=4 cosec 2A . cot 2A</p> <p>= 4(\(\frac 1{sin2A})\) . \(\frac {cos 2A}{sin 2A}\)</p> <p>= \(\frac {4 (cos^2A - sin^2A)}{2 sinA cosA . 2 sinA cosA}\)</p> <p>= \(\frac {cos^2A - sin^2A}{sin^2A cos^2A}\)</p> <p>= \(\frac {cos^2A}{sin^2A cos^2A}\) - \(\frac {sin^2A}{sin^2A cos^2A}\)</p> <p>= \(\frac {1}{sin^2A}\) - \(\frac {1}{cos^2A}\)</p> <p>= cosec<sup>2</sup>A - sec<sup>2</sup>A</p> <p>Hence, L.H.S. = R.H.S. <sub>proved</sub></p> <p></p>
Q18:
Find the value of tan 3A if cosA = \(\frac {15}{17}\).
Type: Short
Difficulty: Easy
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Answer: <p>Here,</p> <p>cosA = \(\frac {15}{17}\)</p> <p>cosA</p> <p>= \(\sqrt {1 - cos^2A}\)</p> <p>= \(\sqrt {1 - (\frac {15}{17}})^2\)</p> <p>= \(\sqrt {1 - \frac {225}{289}}\)</p> <p>= \(\sqrt {\frac {289 - 225}{289}}\)</p> <p>= \(\sqrt {\frac {64}{289}}\)</p> <p>= \(\frac 8{17}\)</p> <p>∴ tanA = \(\frac {sinA}{cosA}\) = \(\frac {\frac {8}{17}}{\frac {15}{17}}\) = \(\frac 8{15}\)</p> <p>Now,</p> <p>tan 3A</p> <p>= \(\frac {3 tanA - tan^3A}{1 - 3 tan^2A}\)</p> <p>= \(\frac {3 × \frac 8{15} - (\frac 8{15})^3}{1 - 3 ×(\frac 8{15})^2}\)</p> <p>= \(\frac {\frac {24}{15} - \frac {512}{3375}}{1 - \frac {192}{225}}\)</p> <p>= \(\frac {\frac {5400 - 512}{3375}}{\frac {225 - 192}{225}}\)</p> <p>= \(\frac {\frac {4888}{3375}}{\frac {33}{225}}\)</p> <p>= \(\frac {4888}{3375}\)× \(\frac {225}{33}\)</p> <p>= \(\frac {4888}{15 × 33}\)</p> <p>= \(\frac {4888}{495}\) <sub>Ans</sub></p> <p></p>
Q19:
Prove that:
\(\frac 1{tan 3A + tanA}\) - \(\frac 1{cot 3A + cotA}\) = cot 4A
Type: Short
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=\(\frac 1{tan 3A + tanA}\) - \(\frac 1{cot 3A + cotA}\)</p> <p>= \(\frac 1{tan 3A + tanA}\) - \(\frac 1{{\frac 1{tan 3A}} + {\frac 1{tanA}}}\)</p> <p>= \(\frac 1{tan 3A + tanA}\) - \(\frac {\frac 1{tanA + tan3A}}{tan3A tanA}\)</p> <p>= \(\frac 1{tan 3A + tanA}\) - \(\frac {tan 3A tanA}{tan 3A + tanA}\)</p> <p>= \(\frac {1 - tan 3A tanA}{tan 3A + tanA}\)</p> <p>= \(\frac {\frac 1{tan 3A + tanA}}{1 - tan3A tanA}\)</p> <p>= \(\frac 1{tan (3A + A)}\)</p> <p>= \(\frac 1{tan 4A}\)</p> <p>= cot 4A</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q20:
Prove that:
\(\frac {1 + sin 2A}{cos 2A}\) = \(\frac {cosA + sinA}{cosA - sinA}\)
Type: Short
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=\(\frac {1 + sin 2A}{cos 2A}\)</p> <p>= \(\frac {sin^2A + cos^2A + 2 sinA cosA}{cos^2A - sin^2A}\)</p> <p>= \(\frac {(cosA + sinA)^2}{(cosA + sinA) (cosA - sinA)}\)</p> <p>= \(\frac {cosA + sinA}{cosA - sinA}\)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q21:
Find the exact value of sin 2\(\theta\) if tan \(\theta\) = \(\frac 43\).
Type: Short
Difficulty: Easy
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Answer: <p>Here,</p> <p>sin 2\(\theta\)</p> <p>= \(\frac {2 tan\theta}{1 + tan^2\theta}\)</p> <p>= \(\frac {2 × \frac 43}{1 + (\frac 43)^2}\)</p> <p>= \(\frac {\frac 83}{1 + \frac {16}{9}}\)</p> <p>= \(\frac {\frac83}{\frac {9 + 16}{9}}\)</p> <p>= \(\frac {\frac 83}{\frac {25}9}\)</p> <p>= \(\frac 83\)× \(\frac 9{25}\)</p> <p>= \(\frac {24}{25}\) <sub>Ans</sub></p>
Q22:
Find the value of sin 2A if cosA = \(\frac 35\).
Type: Short
Difficulty: Easy
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Answer: <p>Here,</p> <p>cos A = \(\frac 35\)</p> <p>sinA</p> <p>= \(\sqrt {1 - cos^2 A}\)</p> <p>= \(\sqrt {1 - (\frac 35)^2}\)</p> <p>= \(\sqrt {1 - \frac 9{25}}\)</p> <p>= \(\sqrt {\frac {25 - 9}{25}}\)</p> <p>= \(\sqrt {\frac {16}{25}}\)</p> <p>= \(\frac 45\)</p> <p>Now,</p> <p>sin 2A</p> <p>= 2 sinA cosA</p> <p>= 2× \(\frac 45\)× \(\frac 35\)</p> <p>= \(\frac {24}{25}\) <sub>Ans</sub></p>
Q23:
Prove that:
sin4A = \(\frac 18\)(3 - 4 cos 2A + cos 4A)
Type: Long
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>= sin<sup>4</sup>A</p> <p>= (sin<sup>2</sup>A)<sup>2</sup></p> <p>=[\(\frac 12\) (1 - cos 2A)]<sup>2</sup></p> <p>= \(\frac 14\) [1 - 2 cos 2A + cos<sup>2</sup>A]</p> <p>= \(\frac14\) [1 - 2 cos2A + (\(\frac {1 + cos4A}{2}\))]</p> <p>= \(\frac 14\) [\(\frac {2 - 4 cos2A + 1 + cos4A}2\)]</p> <p>= \(\frac 18\)(3 - 4 cos 2A + cos 4A)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q24:
Prove that:
cosec 2A + cot 4A = cotA - cosec 4A
Type: Long
Difficulty: Easy
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Answer: <p>Here,</p> <p>cosec 2A + cot 4A = cotA - cosec 4A</p> <p>or, cot 4A + cosec 4A = cotA - cosec 2A</p> <p>L.H.S.</p> <p>= cot 4A + cosec 4A</p> <p>= \(\frac {cos 4A}{sin 4A}\) + \(\frac 1{sin 4A}\)</p> <p>= \(\frac {cos 4A + 1}{sin 4A}\)</p> <p>= \(\frac {2 cos^22A - 1 + 1}{2 sin 2A cos 2A}\)</p> <p>= \(\frac {2 cos^22A}{2 sin 2A cos 2A}\)</p> <p>= \(\frac {cos 2A}{sin 2A}\)</p> <p>= \(\frac {2 cos^2A - 1}{2 sinA cosA}\)</p> <p>= \(\frac {2 cos^2A}{2 sinA cosA}\) - \(\frac 1{2 sinA cosA}\)</p> <p>= cotA - \(\frac 1{sin 2A}\)</p> <p>= cotA - cosec 2A</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q25:
Prove that:
\(\frac {\sqrt 3}{sin 20°}\) - \(\frac 1{cos 20°}\) = 4
Type: Long
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=\(\frac {\sqrt 3}{sin 20°}\) - \(\frac 1{cos 20°}\)</p> <p>= \(\frac {\sqrt 3 cos 20° - sin 20°}{sin 20° cos 20°}\)</p> <p>= \(\frac {2 (\frac {\sqrt 3}{2} cos 20° - \frac 12 sin 20°)}{sin 20° cos 20°}\)</p> <p>= \(\frac {2 × 2 (sin 60° cos 20° - cos 60° sin 20°)}{2 sin 20° cos 20°}\)</p> <p>= \(\frac {4 [sin (60° - 20°)]}{sin 2 × 20°}\)</p> <p>= \(\frac {4 sin 40°}{sin 40°}\)</p> <p>= 4</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q26:
Prove that:
\(\frac 1{sin 10°}\) - \(\frac {\sqrt 3}{cos 10°}\) = 4
Type: Long
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=\(\frac {1}{sin 10°}\) - \(\frac {\sqrt 3}{cos 10°}\)</p> <p>= \(\frac {cos 10° - \sqrt 3 sin 10°}{sin 10° cos 10°}\)</p> <p>= \(\frac {2 (\frac {1}{2} cos 10° - \frac {\sqrt 3}2 sin 10°)}{sin 10° cos 10°}\)</p> <p>= \(\frac {2 × 2 (sin 30° cos10° - cos 30° sin 10°)}{2 sin 10° cos 10°}\)</p> <p>= \(\frac {4 [sin (30° - 10°)]}{sin 2 × 10°}\)</p> <p>= \(\frac {4 sin 20°}{sin 20°}\)</p> <p>= 4</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q27:
Prove that:
cos 5A = 16 cos 5A - 20 cos3A + 5 cosA
Type: Long
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>= cos 5A</p> <p>= cos (2A + 3A)</p> <p>= cos 2A cos 3A - sin 2A sin 3A</p> <p>= (2 cos<sup>2</sup>A - 1) (4 cos<sup>3</sup>A - 3 cosA) - 2 sinA cosA (3 sinA - 4 sin<sup>3</sup>A)</p> <p>= 8 cos<sup>5</sup>A - 6 cos<sup>3</sup>A - 4 cos<sup>3</sup>A + 3 cosA - 6 sin<sup>2</sup>A cosA + 8 sin<sup>4</sup>A cosA</p> <p>= 8 cos<sup>5</sup>A - 10 cos<sup>3</sup>A + 3 cosA - 6 (1 - cos<sup>2</sup>A) cosA + 8 (1 - cos<sup>2</sup>A)<sup>2</sup> cosA</p> <p>= 8 cos<sup>5</sup>A - 10 cos<sup>3</sup>A + 3 cosA - 6 cosA + 6 cos<sup>3</sup>A + 8 cosA (1 - 2 cos<sup>2</sup>A + cos<sup>4</sup>A)</p> <p>= 8 cos<sup>5</sup>A - 4 cos<sup>3</sup>A - 3 cosA + 8 cosA - 16 cos<sup>3</sup>A + 8 cos<sup>5</sup>A</p> <p>=16 cos<sup>5</sup>A - 20 cos<sup>3</sup>A + 5 cosA</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q28:
Prove that:
tan (\(\frac {\pi}{4} + A\)) + tan (\(\frac {\pi}{4} - A\)) = 2 sec 2A
Type: Long
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=tan (\(\frac {\pi}{4} + A\)) + tan(\(\frac {\pi}{4} -A\))</p> <p>= \(\frac {tan \frac {\pi}{4} + tanA}{1 - tan {\frac {\pi}{4}} tanA}\) +\(\frac {tan \frac {\pi}{4} -tanA}{1 +tan {\frac {\pi}{4}} tanA}\)</p> <p>= \(\frac {1 + tanA}{1 - 1 × tanA}\) + \(\frac {1 -tanA}{1 +1 × tanA}\)</p> <p>= \(\frac {(1 + tanA)^2 + (1 - tan A)^2}{(1 - tanA) (1 + tanA)}\)</p> <p>= \(\frac {1 + 2 tanA + tan^2A + 1 - 2 tanA + tan^2A}{1 - tan^2A}\)</p> <p>= \(\frac {2 + 2 tan^2A}{1 - tan^A}\)</p> <p>= \(\frac {2 (1 + tan^2A)}{1 - tan^2A}\)</p> <p>= \(\frac 2{\frac {1 - tan^2A}{1 + tan^2A}}\)</p> <p>= \(\frac 2{cos 2A}\)</p> <p>= 2 sec 2A</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q29:
If tan\(\theta\) = \(\frac 56\) and tan\(\beta\) = \(\frac 1{11}\), show that:
(\(\theta\) + \(\beta\)) = \(\frac {\pi^c}{4}\)
Type: Long
Difficulty: Easy
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Answer: <p>Here,</p> <p>tan\(\theta\) = \(\frac 56\) and tan\(\beta\) = \(\frac 1{11}\)</p> <p>tan (\(\theta\) + \(\beta\)) = \(\frac {tan \theta + tan \beta}{1 - tan\theta tan\beta}\)</p> <p>or,tan (\(\theta\) + \(\beta\)) = \(\frac {\frac 56 + \frac 1{11}}{1 - \frac 56 × \frac 1{11}}\)</p> <p>or, tan (\(\theta\) + \(\beta\)) = \(\frac {\frac {5 × 11 + 1 × 6}{66}}{\frac {66 - 5}{66}}\)</p> <p>or,tan (\(\theta\) + \(\beta\)) = \(\frac {55 + 6}{66}\)× \(\frac {66}{61}\)</p> <p>or,tan (\(\theta\) + \(\beta\)) = \(\frac {61}{66}\)× \(\frac {66}{61}\)</p> <p>or,tan (\(\theta\) + \(\beta\)) = 1</p> <p>or,tan (\(\theta\) + \(\beta\)) = tan 45°</p> <p>or,tan (\(\theta\) + \(\beta\)) = tan\(\frac {\pi^c}{4}\)</p> <p>∴ (\(\theta\) + \(\beta\)) =\(\frac {\pi^c}{4}\)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p> <p></p>
Q30:
Prove that:
cos6\(\theta\) + sin6\(\theta\) = \(\frac 14\)(1 + 3 cos2 2\(\theta\))
Type: Long
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=cos<sup>6</sup>\(\theta\) + sin<sup>6</sup>\(\theta\)</p> <p>= (cos<sup>2</sup>\(\theta\))<sup>3</sup> + (sin<sup>2</sup>\(\theta\))<sup>3</sup></p> <p>= (cos<sup>2</sup>\(\theta\) + sin<sup>2</sup>\(\theta\)) (cos<sup>4</sup>\(\theta\) - cos<sup>2</sup>\(\theta\) sin<sup>2</sup>\(\theta\) + sin<sup>4</sup>\(\theta\))</p> <p>= {(cos<sup>2</sup>\(\theta\))<sup>2</sup> - cos<sup>2</sup>\(\theta\) sin<sup>2</sup>\(\theta\) + (sin<sup>2</sup>\(\theta\))<sup>2</sup>}</p> <p>= {(cos<sup>2</sup>\(\theta\) + sin<sup>2</sup>\(\theta\))<sup>2</sup> - 2 cos<sup>2</sup>\(\theta\) sin<sup>2</sup>\(\theta\) - cos<sup>2</sup>\(\theta\) sin<sup>2</sup>\(\theta\)}</p> <p>= 1 - 3 sin<sup>2</sup>\(\theta\) cos<sup>2</sup>\(\theta\)</p> <p>= 1 - 3 . \(\frac 44\) (sin<sup>2</sup>\(\theta\) cos<sup>2</sup>\(\theta\))</p> <p>= 1 - \(\frac 34\) (4 sin<sup>2</sup>\(\theta\) cos<sup>2</sup>\(\theta\))</p> <p>= 1 - \(\frac 34\) (2 sin\(\theta\) cos\(\theta\))<sup>2</sup></p> <p>= 1 - \(\frac 34\) (sin 2\(\theta\))<sup>2</sup></p> <p>= 1 - \(\frac 34\) (sin<sup>2</sup>2\(\theta\))</p> <p>= 1 - \(\frac 34\) (1 - cos<sup>2</sup>2\(\theta\))</p> <p>= \(\frac {4 - 3 + 3 cos^22\theta}{4}\)</p> <p>= \(\frac {1 + 3 cos^22\theta}{4}\)</p> <p>= \(\frac 14\) (1 + 3 cos<sup>2</sup>2\(\theta\))</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q31:
Prove that:
\(\frac {2 cos 8\theta + 1}{2 cos\theta + 1}\) = (2 cos\(\theta\) - 1) (2 cos 2\(\theta\) - 1) (2 cos 4\(\theta\) - 1)
Type: Long
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=\(\frac {2 cos 8\theta + 1}{2 cos\theta + 1}\)</p> <p>=\(\frac {2 cos 2 × 4\theta + 1}{2 cos\theta + 1}\)</p> <p>=\(\frac {2 (2cos^2 4\theta - 1) + 1}{2cos\theta + 1}\)</p> <p>=\(\frac {4 cos^2 4\theta -2 + 1}{2cos\theta + 1}\)</p> <p>=\(\frac {4 cos^2 4\theta - 1}{2cos\theta + 1}\)</p> <p>=\(\frac {(2cos 4\theta)^2 - (1)^2}{2cos\theta + 1}\)</p> <p>=\(\frac {(2 cos 4\theta + 1)(2 cos 4\theta - 1)}{2cos\theta + 1}\)</p> <p>=\(\frac {(2 cos 2 × 2\theta + 1)(2 cos 4\theta - 1)}{2cos\theta + 1}\)</p> <p>=\(\frac {[2 (2cos^2\theta - 1) + 1](2 cos 4\theta - 1)}{2cos\theta + 1}\)</p> <p>=\(\frac {(4 cos^2\theta - 2 + 1)(2 cos 4\theta - 1)}{2cos\theta + 1}\)</p> <p>=\(\frac {(4 cos^2\theta - 1)(2 cos 4\theta - 1)}{2cos\theta + 1}\)</p> <p>=\(\frac {[(2 cos2\theta)^2 - (1)^2](2 cos 4\theta - 1)}{2cos\theta + 1}\)</p> <p>=\(\frac {(2 cos 2\theta + 1) (2 cos 2\theta - 1) (2 cos 4\theta - 1)}{2cos\theta + 1}\)</p> <p>=\(\frac {[2 × (2 cos^2\theta - 1) + 1] (2 cos 2\theta - 1) (2 cos 4\theta - 1)}{2cos\theta + 1}\)</p> <p>=\(\frac {(4 cos^2\theta - 2 + 1) (2 cos 2\theta - 1) (2 cos 4\theta - 1)}{2cos\theta + 1}\)</p> <p>=\(\frac {(4 cos^2\theta - 1) (2 cos 2\theta - 1) (2 cos 4\theta - 1)}{2cos\theta + 1}\)</p> <p>=\(\frac {[(2 cos\theta)^2 - (1)^2] (2 cos 2\theta - 1) (2 cos 4\theta - 1)}{2cos\theta + 1}\)</p> <p>=\(\frac {(2 cos\theta + 1) (2 cos \theta - 1) (2 cos 2\theta - 1) (2 cos 4\theta - 1)}{2cos\theta + 1}\)</p> <p>=(2 cos\(\theta\) - 1) (2 cos 2\(\theta\) - 1) (2 cos 4\(\theta\) - 1)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p> <p></p> <p></p>
Q32:
Prove that:
cos2A + sin2A cos 2B = cos2B + sin2B cos 2A
Type: Long
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>= cos<sup>2</sup>A + sin<sup>2</sup>A cos 2B</p> <p>= cos<sup>2</sup>A + sin<sup>2</sup>A(1 - 2 sin<sup>2</sup>B)</p> <p>= cos<sup>2</sup>A + sin<sup>2</sup>A - 2 sin<sup>2</sup>A sin<sup>2</sup>B</p> <p>= 1 - 2 sin<sup>2</sup>A sin<sup>2</sup>B</p> <p>= 1 - (1 - cos 2A) sin<sup>2</sup>B</p> <p>= 1 - sin<sup>2</sup>B + sin<sup>2</sup>B cos 2A</p> <p>= cos<sup>2</sup>B + sin<sup>2</sup>B cos 2A</p> <p>Hence, L.H.S. = R.H.S. Proved</p>
Q33:
Express cos 4\(\theta\) in terms of sin\(\theta\).
Type: Long
Difficulty: Easy
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Answer: <p>Here,</p> <p>cos 4\(\theta\)</p> <p>= cos 2 (2\(\theta\))</p> <p>= 1 - 2 sin<sup>2</sup> (2\(\theta\))</p> <p>= 1 - 2 (2sin\(\theta\) cos\(\theta\))<sup>2</sup></p> <p>= 1 - 2× 4 sin<sup>2</sup>\(\theta\) cos<sup>2</sup>\(\theta\)</p> <p>= 1 - 8 sin<sup>2</sup>\(\theta\) (1 - sin<sup>2</sup>\(\theta\))</p> <p>= 1 - 8 sin<sup>2</sup>\(\theta\) + 8 sin<sup>4</sup>\(\theta\) <sub>Ans</sub></p>
Q34:
If tan2\(\alpha\) = 1 + 2 tan2\(\beta\), show that:
cos 2\(\beta\) = 1 + 2 cos 2\(\alpha\)
Type: Long
Difficulty: Easy
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Answer: <p>Here,</p> <p>tan<sup>2</sup>\(\alpha\) = 1 + 2 tan<sup>2</sup>\(\beta\)</p> <p>or, \(\frac {1 - cos 2\alpha}{1 + cos 2\alpha}\) = 1 + 2 (\(\frac {1 - cos 2\beta}{1 + cos 2\beta})\)</p> <p>or,\(\frac {1 - cos 2\alpha}{1 + cos 2\alpha}\) = \(\frac {1 + cos 2\beta + 2 - 2 cos 2\beta}{1 + cos 2\beta}\)</p> <p>or,\(\frac {1 - cos 2\alpha}{1 + cos 2\alpha}\) = \(\frac {3 - cos 2\beta}{1 + cos 2\beta}\)</p> <p>or,\(\frac {1 - cos 2\alpha}{1 + cos 2\alpha}\) + 1 = \(\frac {3 - cos 2\beta}{1 + cos 2\beta}\) + 1</p> <p>or, \(\frac {1 - cos 2\alpha + 1 + cos 2\alpha}{1 + cos 2\alpha}\) = \(\frac {3 - cos 2\beta + 1 + cos 2\beta}{1 + cos 2\beta}\)</p> <p>or, \(\frac 2{1 + cos 2\alpha}\) = \(\frac 4{1 + cos 2\beta}\)</p> <p>or, 2 (1 + cos 2\(\beta\)) = 4 (1 + cos 2\(\alpha\))</p> <p>or, 2 + 2 cos 2\(\beta\) = 4 + 4 cos 2\(\alpha\)</p> <p>or, 2 cos 2\(\beta\) =4 + 4 cos 2\(\alpha\) - 2</p> <p>or,2 cos 2\(\beta\) = 2+ 4 cos 2\(\alpha\)</p> <p>or, cos 2\(\beta\) = \(\frac {2(1 + 2 cos 2\alpha)}{2}\)</p> <p>or,cos 2\(\beta\) =1 + 2 cos 2\(\alpha\)</p> <p>∴ L.H.S. = R.H.S. <sub>Proved</sub></p> <p></p>
Q35:
Show that:
tan\(\theta\) + 2 tan 2\(\theta\) + 4 tan 4\(\theta\) + 8 cot 8\(\theta\) = cot\(\theta\)
Type: Long
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=tan\(\theta\) + 2 tan 2\(\theta\) + 4 tan 4\(\theta\) + 8 cot 8\(\theta\)</p> <p>=tan\(\theta\) + 2 tan 2\(\theta\) + 4 tan 4\(\theta\) + \(\frac 8{tan 8\theta}\)</p> <p>=tan\(\theta\) + 2 tan 2\(\theta\) + 4 tan 4\(\theta\) + \(\frac 8{tan 2⋅4\theta}\)</p> <p>=tan\(\theta\) + 2 tan 2\(\theta\) + 4 tan 4\(\theta\) + \(\frac 8{\frac {2 tan 4\theta}{1 - tan^24\theta}}\)</p> <p>=tan\(\theta\) + 2 tan 2\(\theta\) + 4 tan 4\(\theta\) + \(\frac {8 (1 - tan^24\theta)}{2 tan 4\theta}\)</p> <p>=tan\(\theta\) + 2 tan 2\(\theta\) + \(\frac {4 tan^24\theta + 4 (1 - tan^24\theta)}{tan 4\theta}\)</p> <p>=tan\(\theta\) + 2 tan 2\(\theta\) + \(\frac {4 tan^24\theta + 4 - 4 tan^24\theta}{tan 4\theta}\)</p> <p>=tan\(\theta\) + 2 tan 2\(\theta\) + \(\frac 4{tan 4\theta}\)</p> <p>=tan\(\theta\) + 2 tan 2\(\theta\) + \(\frac 4{tan 2⋅2\theta}\)</p> <p>=tan\(\theta\) + 2 tan 2\(\theta\) + \(\frac 4{\frac {2 tan 2\theta}{1 - tan^22\theta}}\)</p> <p>=tan\(\theta\) + 2 tan 2\(\theta\) + \(\frac {2 (1 - tan^2\theta)}{tan 2\theta}\)</p> <p>= tan\(\theta\) + \(\frac {2 tan^22\theta + 2 - 2 tan^22\theta}{tan 2\theta}\)</p> <p>= tan\(\theta\) + \(\frac 2{tan 2\theta}\)</p> <p>= tan\(\theta\) + \(\frac 2{\frac {2 tan\theta}{1 - tan^2\theta}}\)</p> <p>= tan\(\theta\) + \(\frac {1 - tan^2\theta}{tan\theta}\)</p> <p>= \(\frac {tan^2\theta + 1 - tan^2\theta}{tan\theta}\)</p> <p>= \(\frac 1{tan\theta}\)</p> <p>= cot\(\theta\)</p> <p>Hence, L.H.S. = R.H.S. <sub>proved</sub></p>
Q36:
Show that:
cos 4x - cos 4y = 8 (cos2x - cos2y) (cos2x - sin2y)
Type: Long
Difficulty: Easy
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Answer: <p>R.H.S.</p> <p>= 8 (cos<sup>2</sup>x - cos<sup>2</sup>y) (cos<sup>2</sup>x - sin<sup>2</sup>y)</p> <p>= 2 (2 cos<sup>2</sup>x - 2 cos<sup>2</sup>y) (2 cos<sup>2</sup>x - 2 sin<sup>2</sup>y)</p> <p>= 2 [1 + cos 2x - (2 + cos 2y)] [1 + cos 2x - (1 - cos 2y)]</p> <p>= 2 [1 + cos 2x - 2 - cos 2y] [1 + cos 2x - 1 + cos 2y]</p> <p>= 2 (cos 2x - cos 2y) (cos 2x - cos 2y)</p> <p>= 2 (cos<sup>2</sup>2x - cos<sup>2</sup>2y)</p> <p>= 2 cos<sup>2</sup>2x - 2 cos<sup>2</sup>2y</p> <p>= 1 + cos 4x - (1 + cos 4y)</p> <p>= 1 + cos 4x - 1 - cos 4y</p> <p>= cos 4x - cos 4y</p> <p>Hence, L.H.S. =R.H.S. <sub>Proved</sub></p>
Q37:
Show that:
sec x = \(\frac 2{\sqrt {2 + \sqrt {2 + 2 cos 4x}}}\)
Type: Long
Difficulty: Easy
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Answer: <p>R.H.S.</p> <p>= \(\frac 2{\sqrt {2 + \sqrt {2 + 2 cos 4x}}}\)</p> <p>= \(\frac 2{\sqrt {2 + \sqrt {2 + 2 cos 2 ⋅2x}}}\)</p> <p>=\(\frac 2{\sqrt {2 + \sqrt {2 (1 + cos 2 ⋅2x)}}}\)</p> <p>=\(\frac 2{\sqrt {2 + \sqrt {2 ⋅ 2 cos^22x}}}\)</p> <p>= \(\frac 2{\sqrt {2 + 2 cos 2x}}\)</p> <p>=\(\frac 2{\sqrt {2 (1 + cos 2x)}}\)</p> <p>=\(\frac 2{\sqrt {2 ⋅ 2 cos^2x}}\)</p> <p>= \(\frac 2{2 cosx}\)</p> <p>= \(\frac 1{cosx}\)</p> <p>= sec x</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q38:
Prove that:
4 sin3\(\alpha\) ⋅ cos 3\(\alpha\) + 4 cos3\(\alpha\) ⋅ sin 3\(\alpha\) = 3 sin 4\(\alpha\)
Type: Long
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=4 sin<sup>3</sup>\(\alpha\)⋅ cos 3\(\alpha\) + 4 cos<sup>3</sup>\(\alpha\) ⋅sin 3\(\alpha\)</p> <p>= (3 sin\(\alpha\) - sin3\(\alpha\))⋅ cos 3\(\alpha\) + (3 cos\(\alpha\) + cos 3\(\alpha\))⋅ sin 3\(\alpha\)</p> <p>= 3 sin\(\alpha\) cos 3\(\alpha\) - sin 3\(\alpha\) cos 3\(\alpha\) + 3 cos\(\alpha\) sin 3\(\alpha\) + sin 3\(\alpha\) cos 3\(\alpha\)</p> <p>= 3 sin\(\alpha\) cos 3\(\alpha\) + 3 cos\(\alpha\) sin 3\(\alpha\)</p> <p>= 3 (sin\(\alpha\) cos 3\(\alpha\) + cos\(\alpha\) sin 3\(\alpha\))</p> <p>= 3 sin(\(\alpha\) + 3\(\alpha\))</p> <p>= 3 sin 4\(\alpha\)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q39:
Prove that:
cos3\(\alpha\) . cos 3\(\alpha\) + sin3\(\alpha\) . sin 3\(\alpha\) = cos3 2\(\alpha\)
Type: Long
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=cos<sup>3</sup>\(\alpha\) . cos 3\(\alpha\) + sin<sup>3</sup>\(\alpha\) . sin 3\(\alpha\)</p> <p>= cos<sup>3</sup>\(\alpha\) (4 cos<sup>3</sup>\(\alpha\) - 3 cos\(\alpha\)) + sin<sup>3</sup>\(\alpha\) (3 sin\(\alpha\) - 4 sin<sup>3</sup>\(\alpha\))</p> <p>= 4 cos<sup>6</sup>\(\alpha\) - 3 cos<sup>4</sup>\(\alpha\) + 3 sin<sup>4</sup>\(\alpha\) - 4 sin<sup>6</sup>\(\alpha\)</p> <p>= 4 (cos<sup>6</sup>\(\alpha\) - sin<sup>6</sup>\(\alpha\)) - 3 (cos<sup>4</sup>\(\alpha\) - sin<sup>4</sup>\(\alpha\))</p> <p>= 4 {(cos<sup>2</sup>\(\alpha\))<sup>3</sup> - (sin<sup>2</sup>\(\alpha\))<sup>3</sup>} - 3{(cos<sup>2</sup>\(\alpha\))<sup>2</sup> - (sin<sup>2</sup>\(\alpha\))<sup>2</sup>}</p> <p>= 4 (cos<sup>2</sup>\(\alpha\) - sin<sup>2</sup>\(\alpha\)) (cos<sup>4</sup>\(\alpha\) + cos<sup>2</sup>\(\alpha\) sin<sup>2</sup>\(\alpha\) + sin<sup>4</sup>\(\alpha\)) - 3 (cos<sup>2</sup>\(\alpha\) +sin<sup>2</sup>\(\alpha\))(cos<sup>2</sup>\(\alpha\) - sin<sup>2</sup>\(\alpha\))</p> <p>= 4 cos 2\(\alpha\) {(cos<sup>2</sup>\(\alpha\))<sup>2</sup> + (sin<sup>2</sup>\(\alpha\))<sup>2</sup> + cos<sup>2</sup>\(\alpha\) sin<sup>2</sup>\(\alpha\)} - 3 cos 2\(\alpha\)</p> <p>= cos 2\(\alpha\) [4 {(cos<sup>2</sup>\(\alpha\) + sin<sup>2</sup>\(\alpha\))<sup>2</sup> - 2 cos<sup>2</sup>\(\alpha\) sin<sup>2</sup>\(\alpha\) + cos<sup>2</sup>\(\alpha\) sin<sup>2</sup>\(\alpha\)} - 3]</p> <p>= cos 2\(\alpha\) [4 - 4 cos<sup>2</sup>\(\alpha\) sin<sup>2</sup>\(\alpha\) - 3]</p> <p>= cos 2\(\alpha\) [1 - sin<sup>2</sup>2\(\alpha\)]</p> <p>= cos 2\(\alpha\)⋅ cos<sup>2</sup>2\(\alpha\)</p> <p>= cos<sup>3</sup>2\(\alpha\)</p> <p>∴ L.H.S. = R.H.S. <sub>Proved</sub></p>
Q40:
Prove that:
(2 cos\(\theta\) + 1) (2 cos\(\theta\) - 1) (2 cos 2\(\theta\) - 1) = 2 (cos 4\(\theta\) + 1)
Type: Long
Difficulty: Easy
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Answer: <p>R.H.S.</p> <p>=2 (cos 4\(\theta\) + 1)</p> <p>= 2 [cos 2. 2\(\theta\)] + 1</p> <p>= 2 [2 cos<sup>2</sup> 2\(\theta\) - 1] + 1</p> <p>= 4 cos<sup>2</sup> 2\(\theta\) - 2 + 1</p> <p>= 4 (2 cos<sup>2</sup>\(\theta\) - 1)<sup>2</sup> - 1</p> <p>= 4 (4 cos<sup>4</sup>\(\theta\) - 4 cos<sup>2</sup>\(\theta\) + 1) - 1</p> <p>= 16 cos<sup>4</sup>\(\theta\) - 16 cos<sup>2</sup>\(\theta\) + 4 - 1</p> <p>= 16 cos<sup>4</sup>\(\theta\) - 16 cos<sup>2</sup>\9\theta\) + 3</p> <p>= 16 cos<sup>4</sup>\(\theta\) - 12 cos<sup>2</sup>\(\theta\) - 4 cos<sup>2</sup>\(\theta\) + 3</p> <p>= 4 cos<sup>2</sup>\(\theta\) (4 cos<sup>2</sup>\(\theta\) - 3) - 1 (4 cos<sup>2</sup>\(\theta\) - 3)</p> <p>= (4 cos<sup>2</sup>\(\theta\) - 3) (4 cos<sup>2</sup>\(\theta\) - 1)</p> <p>=(4 cos<sup>2</sup>\(\theta\) - 2 - 1) [(2 cos\(\theta\))<sup>2</sup> - (1)<sup>2</sup>]</p> <p>= [2 (2 cos<sup>2</sup>\(\theta\) -1) - 1] (2 cos\(\theta\) + 1) (2 cos\(\theta\) - 1)</p> <p>= (2 cos 2\(\theta\) - 1)(2 cos\(\theta\) + 1) (2 cos\(\theta\) - 1)</p> <p>=(2 cos\(\theta\) + 1) (2 cos\(\theta\) - 1)(2 cos 2\(\theta\) - 1)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q41:
Prove that:
2 (cos6x + sin6x) - 3 (cos4x + sin4x) + 1 = 0
Type: Long
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=2 (cos<sup>6</sup>x + sin<sup>6</sup>x) - 3 (cos<sup>4</sup>x + sin<sup>4</sup>x) + 1</p> <p>= 2 [(cos<sup>2</sup>x)<sup>3</sup> + (sin<sup>2</sup>x)<sup>3</sup>] - 3 [(cos<sup>2</sup>x)<sup>2</sup> + (sin<sup>2</sup>x)<sup>2</sup>] + 1</p> <p>= 2 (cos<sup>2</sup>x + sin<sup>2</sup>x) [(cos<sup>2</sup>x)<sup>2</sup> - cos<sup>2</sup>x sin<sup>2</sup>x + (sin<sup>2</sup>x)<sup>2</sup>] - 3 [(cos<sup>2</sup>x + sin<sup>2</sup>x)<sup>2</sup> - 2 cos<sup>2</sup>x sin<sup>2</sup>x] + 1</p> <p>= 2× 1 [(cos<sup>2</sup>x + sin<sup>2</sup>x)<sup>2</sup> - 2 cos<sup>2</sup>x sin<sup>2</sup>x - cos<sup>2</sup>x sin<sup>2</sup>x] - 3 [1 - 2 sin<sup>2</sup>x cos<sup>2</sup>x] + 1</p> <p>= 2 (1 - 3 cos<sup>2</sup>x sin<sup>2</sup>x) - 3 + 6 cos<sup>2</sup>x sin<sup>2</sup>x + 1</p> <p>= 2 - 6 cos<sup>2</sup>x sin<sup>2</sup>x - 2 +6 cos<sup>2</sup>x sin<sup>2</sup>x</p> <p>= 0</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q42:
Prove that:
\(\frac {sin^2A - sin^2B}{sinA cosA - sinB cosB}\) = tan (A + B)
Type: Long
Difficulty: Easy
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Answer: <p>R.H.S.</p> <p>= tan (A + B)</p> <p>= \(\frac {sin (A + B)}{cos (A + B)}\)× \(\frac {sin (A - B)}{sin (A - B)}\)</p> <p>= \(\frac {(sinA cosB + cosA sinB) (sinA cosB - cosA sinB)}{(cosA cosB - sinA sinB) (sinA cosB - cosA sinB)}\)</p> <p>= \(\frac {sin^2A cos^2B - cos^2A sin^2B}{sinA cosA cos^2B - sinB cosB cos^2A - sinB cosB sin^2A - cosA sinA sin^2B}\)</p> <p>= \(\frac {sin^2A (1 - sin^2B) - (1 - sin^2A) sin^2B}{sinA cosA (cos^2B + sin^2B) - sinB cosB (cos^2A + sin^2A)}\)</p> <p>= \(\frac {sin^2A - sin^2A sin^2B - sin^2B + sin^2A sin^2B}{sinA cosA . 1 - sinB cosB . 1}\)</p> <p>=\(\frac {sin^2A - sin^2B}{sinA cosA - sinB cosB}\)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p> <p></p>
Q43:
Prove that:
cos6\(\theta\) - sin6\(\theta\) = cos 2\(\theta\) (1 - \(\frac 14\) sin2 2\(\theta\))
Type: Long
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=cos<sup>6</sup>\(\theta\) - sin<sup>6</sup>\(\theta\)</p> <p>= (cos<sup>2</sup>\(\theta\))<sup>3</sup>- (sin<sup>2</sup>\(\theta\))<sup>3</sup></p> <p>= (cos<sup>2</sup>\(\theta\) - sin<sup>2</sup>\(\theta\)) [(cos<sup>2</sup>\(\theta\))<sup>2</sup> + cos<sup>2</sup>\(\theta\) sin<sup>2</sup>\(\theta\) + (sin<sup>2</sup>\(\theta\))<sup>2</sup>]</p> <p>= cos 2\(\theta\) [(cos<sup>2</sup>\(\theta\))<sup>2</sup> + 2 cos<sup>2</sup>\(\theta\) sin<sup>2</sup>\(\theta\) + (sin<sup>2</sup>\(\theta\))<sup>2</sup> - cos<sup>2</sup>\(\theta\) sin<sup>2</sup>\(\theta\)]</p> <p>= cos 2\(\theta\) [(cos<sup>2</sup>\(\theta\) + sin<sup>2</sup>\(\theta\))<sup>2</sup> - \(\frac 14\) × 4 cos<sup>2</sup>\(\theta\) sin<sup>2</sup>\(\theta\)]</p> <p>= cos 2\(\theta\) [(1)<sup>2</sup> - \(\frac 14\) (2 sin\(\theta\) cos\(\theta\))<sup>2</sup>]</p> <p>= cos 2\(\theta\) [1 - \(\frac 14\) (sin 2\(\theta\))<sup>2</sup>]</p> <p>=cos 2\(\theta\) (1 - \(\frac 14\) sin<sup>2</sup> 2\(\theta\))</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q44:
$$ \frac{1-2sinA.cosA}{2}=sin^2(\frac{\pi^c}{4}-A)$$
Type: Short
Difficulty: Easy
Q45:
$$ \frac{2sin\:\beta=sin2\beta}{2sin\beta-sin2\beta}=cot^2 \: \frac{\beta}{2}$$
Type: Short
Difficulty: Easy
Q46:
$$ \frac{cos\; \alpha}{1-sin\alpha}=\frac{1+tan\: \alpha/2}{1-tan\:\alpha/2}$$
Type: Short
Difficulty: Easy
Q47:
$$cot\:A=\frac{1}{2} (cot\frac{A}{2}-tan\frac{A}{2}$$
Type: Short
Difficulty: Easy
Q48:
$$cos^2(\frac{\pi}{4}-\frac{\theta}{4})-sin^2(\frac{\pi}{4}-\frac{\theta}{4})=sin\frac{\theta}{2}$$
Type: Short
Difficulty: Easy
Q49:
$$ \frac{1-tan^2\:(\frac{\pi}{4}-\frac{\theta}{4}}{1+tan^2(\frac{\pi}{4}-\frac{\theta}{4})}=sin\frac{\theta}{2}$$
Type: Short
Difficulty: Easy
Q50:
$$\frac{sin\frac{\theta}{2}-\sqrt{1+ sin\theta}}{\cos\frac{\theta}{2}-\sqrt{1+sin\theta}}=cot\frac{\theta}{2}$$
Type: Short
Difficulty: Easy
Q51:
$$\frac{\sin\frac{A}{2}+sinA}{1+cos\frac{A}{2}+cosA}=\tan\frac{A}{2}$$
Type: Short
Difficulty: Easy
Q52:
$$\tan\frac{\theta}{2}=\frac{1+sin\theta-cos\theta}{1+sin\theta+cos\theta}$$
Type: Short
Difficulty: Easy
Q53:
$$\frac{1+cos\theta+sin\theta}{1-cos\theta+sin\theta}=cot\frac{\theta}{2}$$
Type: Short
Difficulty: Easy
Q54:
$$\frac{1+cosA+sinA}{1-cosA+sinA}=\frac{1}{tan\frac{A}{2}}$$
Type: Short
Difficulty: Easy
Q55:
$$\frac{1+cos\theta}{sin\theta}=\cot\frac{\theta}{2}$$
Type: Short
Difficulty: Easy
Q56:
$$\cot\frac{A}{2}-tan\frac{A}{2}=2\:cot\:A$$
Type: Short
Difficulty: Easy
