Subjective Questions
Q1:
If tan\(\frac {\theta}{2}\) = \(\frac 34\), prove that:
cos\(\theta\) = \(\frac 7{25}\)
Type: Short
Difficulty: Easy
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Answer: <p>Here,</p> <p>tan\(\frac {\theta}{2}\) = \(\frac 34\)</p> <p>L.H.S.</p> <p>= cos\(\theta\)</p> <p>= \(\frac {1 - tan^2\frac {\theta}2}{1 + tan^2\frac {\theta}2}\)</p> <p>= \(\frac {1 - (\frac 34)^2}{1 + (\frac 34)^2}\)</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>= \(\frac 7{25}\)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q2:
If cos\(\frac {\theta}{2}\) = \(\frac 23\), prove that:
cos\(\theta\) = \(\frac {-22}{27}\)
Type: Short
Difficulty: Easy
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Answer: <p>Here,</p> <p>cos\(\frac {\theta}{2}\) = \(\frac 23\)</p> <p>L.H.S.</p> <p>=cos\(\theta\)</p> <p>= 4 cos<sup>3</sup>\(\frac {\theta}3\) - 3 cos\(\frac {\theta}3\)</p> <p>= 4× (\(\frac 23\))<sup>3</sup> - 3× \(\frac 23\)</p> <p>= 4× \(\frac 8{27}\) - \(\frac 63\)</p> <p>= \(\frac {32 - 54}{27}\)</p> <p>= \(\frac {-22}{27}\)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q3:
If tan\(\frac {\alpha}3\) = \(\frac 15\), prove that:
tan\(\alpha\) = \(\frac {37}{55}\)
Type: Short
Difficulty: Easy
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Answer: <p>Here,</p> <p>tan\(\frac {\alpha}3\) = \(\frac 15\)</p> <p>L.H.S.</p> <p>=tan\(\alpha\)</p> <p>= \(\frac {3 tan{\frac {\alpha}3} - tan^3{\frac {\alpha}3}}{1 - 3 tan^2 {\frac {\alpha}3}}\)</p> <p>= \(\frac {3 × {\frac 15} - (\frac 15)^3}{1 - 3 × (\frac 15)^2}\)</p> <p>= \(\frac {\frac 35 - \frac 1{125}}{1 - \frac 3{25}}\)</p> <p>= \(\frac {\frac {75 - 1}{125}}{\frac {25 - 3}{25}}\)</p> <p>= \(\frac {74}{125}\)× \(\frac {25}{25}\)</p> <p>= \(\frac {37}{55}\)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q4:
Prove that:
\(\frac {1 + cos\theta}{1 - cos\theta}\) = cot2\(\frac {\theta}2\)
Type: Short
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=\(\frac {1 + cos\theta}{1 - cos\theta}\)</p> <p>= \(\frac {1 + 2 cos^2{\frac \theta2} - 1}{1 - (1 - 2sin^2\frac {\theta}2)}\)</p> <p>=\(\frac {2 cos^2{\frac \theta2}}{1 - 1 + 2sin^2\frac {\theta}2}\)</p> <p>= \(\frac {2 cos^2\frac \theta2}{2 sin^2\frac \theta2}\)</p> <p>=\(\frac {cos^2\frac \theta2}{sin^2\frac \theta2}\)</p> <p>= cot<sup>2</sup>\(\frac \theta2\)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p> <p></p>
Q5:
Prove that:
\(\frac {1 - tan^2({\frac \pi4} - {\frac \theta4})}{1 + tan^2({\frac \pi4} - {\frac \theta4})}\) = sin\(\frac \theta2\)
Type: Short
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=\(\frac {1 - tan^2({\frac \pi4} - {\frac \theta4})}{1 +tan^2({\frac \pi4} - {\frac \theta4})}\)</p> <p>= cos 2(\(\frac \pi4\) - \(\frac \theta4\))</p> <p>= cos (\(\frac \pi2\) - \(\frac \theta2\))</p> <p>= sin\(\frac \theta2\)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p> <p></p> <p></p>
Q6:
Prove that:
\(\frac {1 + sin\theta - cos\theta}{1 + sin\theta + cos\theta}\) = tan\(\frac \theta2\)
Type: Short
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=\(\frac {1 + sin\theta - cos\theta}{1 + sin\theta + cos\theta}\)</p> <p>= \(\frac {1 + 2 sin\frac \theta2 cos\frac \theta2 - 1 + 2 sin^2\frac \theta2}{1 + 2 sin\frac \theta2 cos\frac \theta2 + 2 cos^2\frac \theta2 - 1}\)</p> <p>= \(\frac {2 sin\frac \theta2 (cos\frac \theta2 + sin\frac \theta2)}{2 cos\frac \theta2 (cos\frac \theta2 + sin\frac \theta2)}\)</p> <p>= \(\frac {sin\frac \theta2}{cos\frac \theta2}\)</p> <p>= tan\(\frac \theta2\)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q7:
Prove that:
\(\frac {sin^3 \frac \theta2 + cos^3 \frac \theta2}{sin \frac \theta2 + cos \frac \theta2}\) = 1 - \(\frac 12\)sin\(\theta\)
Type: Short
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=\(\frac {sin^3 \frac \theta2 + cos^3 \frac \theta2}{sin \frac \theta2 + cos \frac \theta2}\)</p> <p>= \(\frac {(sin \frac \theta2 + cos \frac \theta2) (sin^2\frac \theta2 + cos^2\frac \theta2 - sin\frac \theta2 cos\frac \theta2)}{(sin\frac \theta2 + cos\frac \theta2)}\)</p> <p>= 1 - \(\frac 22\) sin\(\frac \theta2\) cos\(\frac \theta2\)</p> <p>= 1 - \(\frac 12\)sin\(\theta\)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q8:
Prove that:
\(\frac {sin\frac A2 + sinA}{1 + cos\frac A2 + cosA}\) = tan \(\frac A2\)
Type: Short
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=\(\frac {sin\frac A2 + sinA}{1 + cos\frac A2 + cosA}\)</p> <p>= \(\frac {sin\frac A2 + 2 sin\frac A2 cos\frac A2}{1 + cos\frac A2 + 2 cos^2\frac A2 - 1}\)</p> <p>= \(\frac {sin \frac A2 (1 + 2 cos\frac A2)}{cos \frac A2 (1 + 2 cos\frac A2)}\)</p> <p>= tan\(\frac A2\)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p> <p></p>
Q9:
If sin \(\frac \theta3\) = \(\frac 12\), find the value of sin\(\theta\).
Type: Short
Difficulty: Easy
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Answer: <p>Here,</p> <p>sin \(\frac \theta3\) = \(\frac 12\)</p> <p>sin\(\theta\)</p> <p>= 3 sin \(\frac \theta3\) - 4 sin<sup>3</sup>\(\frac \theta3\)</p> <p>= 3× \(\frac 12\) - 4× (\(\frac 12\))<sup>3</sup></p> <p>= \(\frac 32\) - \(\frac 12\)</p> <p>= \(\frac {3 - 1}{2}\)</p> <p>= \(\frac 22\)</p> <p>= 1 <sub>Ans</sub></p>
Q10:
Prove that:
\(\frac {cos\alpha}{1 - sin\alpha}\) = \(\frac {1 + tan\frac \alpha2}{1 - tan\frac \alpha2}\)
Type: Short
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=\(\frac {cos\alpha}{1 - sin\alpha}\)</p> <p>= \(\frac {\frac {1 - tan^2\frac \alpha2}{1 + tan^2\frac \alpha2}}{1 - \frac {2 tan\frac \alpha2}{1 + tan^2\frac \alpha2}}\)</p> <p>=\(\frac {\frac {1 - tan^2\frac \alpha2}{1 + tan^2\frac \alpha2}}{\frac {1 + tan^2\frac \alpha2 - 2 tan\frac \alpha2}{1 + tan^2\frac \alpha2}}\)</p> <p>= \(\frac {1 - tan^2\frac \alpha2}{(1 - tan\frac \alpha2)^2}\)</p> <p>= \(\frac {(1 + tan\frac \alpha2)(1 - tan\frac \alpha2)}{(1 - tan\frac \alpha2)^2}\)</p> <p>=\(\frac {1 + tan\frac \alpha2}{1 - tan\frac \alpha2}\)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q11:
Prove that:
\(\frac {sin 2A}{1 + cos 2A}\) × \(\frac {cosA}{1 + cosA}\) = tan\(\frac A2\)
Type: Short
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=\(\frac {sin 2A}{1 + cos 2A}\)× \(\frac {cosA}{1 + cosA}\)</p> <p>= \(\frac {2 sinA cosA}{2 cos^2A}\)× \(\frac {cosA}{1 + cosA}\)</p> <p>= \(\frac {sinA}{1 + cosA}\)</p> <p>= \(\frac {2 sin\frac A2 cos\frac A2}{2 cos^2\frac A2}\)</p> <p>= tan\(\frac A2\)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p> <p></p>
Q12:
If cos 30° = \(\frac {\sqrt 3}2\), find the value of sin 15°.
Type: Short
Difficulty: Easy
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Answer: <p>Let A = 30°</p> <p>cos 30° = \(\frac {\sqrt 3}2\)</p> <p>We know,</p> <p>sin \(\frac A2\) =± \(\sqrt {\frac {1 - cosA}{2}}\)</p> <p>sin \(\frac {30°}2\) =± \(\sqrt {\frac {1 - cos 30°}{2}}\)</p> <p>sin 15° =± \(\sqrt {\frac {1 - \frac {\sqrt 3}2}{\frac 21}}\)</p> <p>or, sin 15° = ± \(\sqrt {\frac {2 - \sqrt 3}{2} × \frac {1}{2}}\)</p> <p>or, sin 15° =± \(\sqrt {\frac {2 - \sqrt 3}{4}}\)</p> <p>or, sin 15° =± \(\sqrt {\frac {4 - 2\sqrt 3}{8}}\)</p> <p>or, sin 15° =± \(\sqrt {\frac {3 - 2\sqrt 3 + 1}{8}}\)</p> <p>or, sin 15° =± \(\sqrt {\frac {(\sqrt 3 -1)^2}{8}}\)</p> <p>∴ sin 15° =± \(\frac {\sqrt 3 -1}{2\sqrt 2}\) <sub>Ans</sub></p>
Q13:
Prove that:
\(\frac {2 sin\beta + sin 2\beta}{2 sin\beta - sin 2\beta}\) = cot2\(\frac \beta2\)
Type: Short
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=\(\frac {2 sin\beta + sin 2\beta}{2 sin\beta - sin 2\beta}\)</p> <p>=\(\frac {2 sin\beta + 2 sin\beta cos\beta}{2 sin\beta - 2 sin\beta cos\beta}\)</p> <p>= \(\frac {2 sin\beta (1 + cos\beta)}{2 sin\beta (1 - cos\beta)}\)</p> <p>= \(\frac {2 cos^2\frac \beta2}{2 sin^2\frac \beta2}\)</p> <p>[\(\because\) 1 + cos\(\theta\) = 2 cos<sup>2</sup>\(\frac \theta2\),1 - cos\(\theta\) = 2 sin<sup>2</sup>\(\frac \theta2\)]</p> <p>= cot<sup>2</sup>\(\frac \beta2\)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q14:
If sin \(\frac \theta3\) = \(\frac 45\), find the value of sin\(\theta\).
Type: Short
Difficulty: Easy
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Answer: <p>Here,</p> <p>sin \(\frac \theta3\) = \(\frac 45\)</p> <p>sin\(\theta\)</p> <p>= 3 sin \(\frac \theta3\) - 4 sin<sup>3</sup>\(\frac \theta3\)</p> <p>= 3× \(\frac 45\) - 4× (\(\frac 45\))<sup>3</sup></p> <p>= \(\frac {12}5\) - \(\frac {256}{125}\)</p> <p>= \(\frac {300 - 256}{125}\)</p> <p>= \(\frac {44}{125}\) <sub>Ans</sub></p>
Q15:
If cos\(\frac \theta3\) = \(\frac 35\), find the value of cos\(\theta\).
Type: Short
Difficulty: Easy
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Answer: <p>Here,</p> <p>cos\(\frac \theta3\) = \(\frac 35\)</p> <p>cos \(\theta\)</p> <p>= 4 cos<sup>3</sup>\(\frac \theta3\) - 3 cos\(\frac \theta3\)</p> <p>= 4× (\(\frac 35\))<sup>3</sup>- 3× \(\frac 35\)</p> <p>= 4× \(\frac {27}{125}\) - \(\frac 95\)</p> <p>= \(\frac {108 - 225}{125}\)</p> <p>= -\(\frac {117}{125}\) <sub>Ans</sub></p>
Q16:
Find the value of sin\(\theta\) if sin\(\frac \theta2\) = \(\frac 35\).
Type: Short
Difficulty: Easy
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Answer: <p>Here,</p> <p>sin\(\frac \theta2\) = \(\frac 35\)</p> <p>cos\(\frac \theta2\)</p> <p>= \(\sqrt {1 - sin^2\frac \theta2}\)</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\(\theta\) = 2 sin\(\frac \theta2\)⋅ cos\(\frac \theta2\) = 2× \(\frac 35\)× \(\frac 45\) = \(\frac {24}{25}\) <sub>Ans</sub></p>
Q17:
Prove that:
\(\frac {tanx + sinx}{2 tanx}\) = cos2\(\frac x2\)
Type: Short
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=\(\frac {tanx + sinx}{2 tanx}\)</p> <p>= \(\frac {\frac {sinx}{cosx} + sinx}{\frac {2 sinx}{cosx}}\)</p> <p>= \(\frac {\frac {sinx + sinx cosx}{cosx}}{\frac {2sinx}{cosx}}\)</p> <p>= \(\frac {sinx (1 + cosx)}{2 sinx}\)</p> <p>= \(\frac {1 + cosx}{2}\)</p> <p>= cos<sup>2</sup>\(\frac x2\)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q18:
Prove that:
\(\frac {2 sinA - sin 2A}{2 sinA + sin2A}\) = tan2\(\frac A2\)
Type: Short
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=\(\frac {2 sinA - sin 2A}{2 sinA + sin2A}\)</p> <p>= \(\frac {2 sinA - 2 sinA cosA}{2 sinA + 2 sinA cosA}\)</p> <p>= \(\frac {2 sinA (1 - cosA)}{2 sinA (1 + cosA)}\)</p> <p>= \(\frac {1 - cosA}{1 + cosA}\)</p> <p>= \(\frac {2 sin^2\frac A2}{2 cos^2\frac A2}\)</p> <p>= tan<sup>2</sup>\(\frac A2\)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q19:
If sin 45° = \(\frac 1{\sqrt2}\), find the value of sin22\(\frac 12\)°.
Type: Short
Difficulty: Easy
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Answer: <p>Here,</p> <p>sin 45° = \(\frac 1{\sqrt2}\)</p> <p>We know,</p> <p>cos45°</p> <p>= \(\sqrt {1 - sin^245°}\)</p> <p>= \(\sqrt {1 - (\frac 1{\sqrt 2})^2}\)</p> <p>= \(\sqrt {1 - \frac 12}\)</p> <p>= \(\frac 1{\sqrt 2}\)</p> <p>Now,</p> <p>cosA= 1 - 2 sin<sup>2</sup>\(\frac A2\)</p> <p>Let: A = 45°</p> <p>or, cos 45° = 1 - 2 sin<sup>2</sup>22\(\frac 12\)°</p> <p>or, 2 sin<sup>2</sup>22\(\frac 12\)° = 1 - cos 45°</p> <p>or,2 sin<sup>2</sup>22\(\frac 12\)° = 1 - \(\frac 1{\sqrt 2}\)</p> <p>or,2 sin<sup>2</sup>22\(\frac 12\)° = \(\frac {\sqrt 2 -1}{\sqrt 2}\)</p> <p>or,2 sin<sup>2</sup>22\(\frac 12\)° = \(\frac {\sqrt 2 - 1}{\sqrt 2}\)× \(\frac {\sqrt 2}{\sqrt 2}\)</p> <p>or,2 sin<sup>2</sup>22\(\frac 12\)° = \(\frac {2 - \sqrt 2}{2 × 2}\)</p> <p>or,2 sin<sup>2</sup>22\(\frac 12\)° = \(\frac {2 - \sqrt 2}{4}\)</p> <p>or,2 sin<sup>2</sup>22\(\frac 12\)° =± \(\sqrt {\frac {2 - \sqrt 2}{4}}\)</p> <p>∴2 sin<sup>2</sup>22\(\frac 12\)° =± \(\frac 12\) \(\sqrt {2 - \sqrt 2}\) <sub>Ans</sub></p>
Q20:
Prove that:
cos2(\(\frac \pi4\) - \(\frac \theta4\)) - sin2(\(\frac \pi4\) - \(\frac \theta4\)) = sin\(\frac \theta2\)
Type: Short
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=cos<sup>2</sup>(\(\frac \pi4\) - \(\frac \theta4\)) - sin<sup>2</sup>(\(\frac \pi4\) - \(\frac \theta4\))</p> <p>= cos 2(\(\frac \pi4\) - \(\frac \theta4\)) [\(\because\) cos<sup>2</sup>A - sin<sup>2</sup>A = cos 2A]</p> <p>= cos(\(\frac \pi2\) - \(\frac \theta2\))</p> <p>= cos (90 - \(\frac \theta2\))</p> <p>= sin\(\frac \theta2\)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q21:
If cos\(\frac \theta3\) = \(\frac 12\)(a + \(\frac 1a\)), prove that:
cos\(\theta\) = \(\frac 12\)(a3 + \(\frac 1{a^3}\))
Type: Short
Difficulty: Easy
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Answer: <p>Here,</p> <p>cos\(\frac \theta3\) = \(\frac 12\)(a + \(\frac 1a\))</p> <p>L.H.S.</p> <p>=cos\(\theta\)</p> <p>= 4 cos<sup>3</sup>\(\frac \theta3\) - 3 cos\(\frac \theta3\)</p> <p>= 4 [\(\frac 12\)(a + \(\frac 1a\))]<sup>3</sup> - 3 [\(\frac 12\)(a + \(\frac 1a\))]</p> <p>= 4× \(\frac 18\) (a + \(\frac 1a\))<sup>3</sup> - 3×\(\frac 12\)(a + \(\frac 1a\))</p> <p>= \(\frac 12\)[(a + \(\frac 1a\))<sup>3</sup> - 3× (a + \(\frac 1a\))]</p> <p>= \(\frac 12\) [a<sup>3</sup> + \(\frac 1{a^3}\) + 3.a.\(\frac 1a\) - 3(a + \(\frac 1a\))]</p> <p>= \(\frac 12\)(a<sup>3</sup> + \(\frac 1{a^3}\))</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q22:
Prove that:
cotA = \(\frac 12\)(cot\(\frac A2\) - tan\(\frac A2\))
Type: Short
Difficulty: Easy
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Answer: <p>R.H.S.</p> <p>= \(\frac 12\)(cot\(\frac A2\) - tan\(\frac A2\))</p> <p>= \(\frac 12\)(\(\frac {cos\frac A2}{sin \frac A2}\) - \(\frac {sin\frac A2}{cos\frac A2}\))</p> <p>= \(\frac 12\)(\(\frac {cos^2\frac A2 - sin^2\frac A2}{sin\frac A2 cos\frac A2}\))</p> <p>= \(\frac {cosA}{sinA}\)</p> <p>= cot A <sub>Proved</sub></p>
Q23:
If cos\(\frac A3\) = \(\frac 12\)(p + \(\frac 1p\)), prove that:
cosA = \(\frac 12\)(p3 + \(\frac 1{p^3}\))
Type: Short
Difficulty: Easy
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Answer: <p>Here,</p> <p>cos\(\frac A3\) = \(\frac 12\)(p + \(\frac 1p\))</p> <p>L.H.S.</p> <p>=cosA</p> <p>= 4 cos<sup>3</sup>\(\frac A3\) - 3 cos\(\frac A3\)</p> <p>= 4 [\(\frac 12\)(p + \(\frac 1p\))]<sup>3</sup> - 3 [\(\frac 12\)(p + \(\frac 1p\))]</p> <p>= 4× \(\frac 18\) (p + \(\frac 1p\))<sup>3</sup> - 3×\(\frac 12\)(p + \(\frac 1p\))</p> <p>= \(\frac 12\)[(p + \(\frac 1p\))<sup>3</sup> - 3× (p + \(\frac 1p\))]</p> <p>= \(\frac 12\) [p<sup>3</sup> + \(\frac 1{p^3}\) + 3.p.\(\frac 1p\) - 3(p + \(\frac 1p\))]</p> <p>= \(\frac 12\)(p<sup>3</sup> + \(\frac 1{p^3}\))</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q24:
If sin\(\frac \alpha3\) = \(\frac 35\), find the value of sin\(\alpha\).
Type: Short
Difficulty: Easy
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Answer: <p>Here,</p> <p>sin\(\frac \alpha3\) = \(\frac 35\)</p> <p>sin \(\alpha\)</p> <p>= 3 sin\(\frac \alpha3\) - 4 sin<sup>3</sup>\(\frac \alpha3\)</p> <p>= 3 × \(\frac 35\) - 4 × (\(\frac 35\))<sup>3</sup></p> <p>= \(\frac 95\) - 4 × \(\frac {27}{125}\)</p> <p>= \(\frac {225 - 108}{125}\)</p> <p>= \(\frac {117}{125}\) <sub>Ans</sub></p>
Q25:
If sin\(\alpha\) = \(\frac 35\), find the values of cos 2\(\alpha\) and sin 3\(\alpha\).
Type: Short
Difficulty: Easy
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Answer: <p>Here,</p> <p>sin\(\alpha\) = \(\frac 35\)</p> <p>cos\(\alpha\) = \(\sqrt {1 - sin^2\alpha}\) = \(\sqrt {1 - (\frac 35)^2}\) = \(\sqrt {1 - \frac 9{25}}\) = \(\sqrt {\frac {25 - 9}{25}}\) = \(\sqrt {\frac {16}{25}}\) = \(\frac 45\)</p> <p>Now,</p> <p>cos 2\(\alpha\)</p> <p>= cos<sup>2</sup>\(\alpha\) - sin<sup>2</sup>\(\alpha\)</p> <p>= (\(\frac 45\))<sup>2</sup>- (\(\frac 35\))<sup>2</sup></p> <p>= \(\frac {16}{25}\) - \(\frac {9}{25}\)</p> <p>= \(\frac {16 - 9}{25}\)</p> <p>= \(\frac 7{25}\)</p> <p>Again,</p> <p>sin 3\(\alpha\)</p> <p>= 3 sin\(\alpha\) - 4 sin<sup>3</sup>\(\alpha\)</p> <p>= 3× \(\frac 35\) - 4× (\(\frac 35\))<sup>3</sup></p> <p>= \(\frac 95\) - 4 × \(\frac {27}{125}\)</p> <p>= \(\frac 95\) - \(\frac {108}{125}\)</p> <p>= \(\frac {225 - 108}{125}\)</p> <p>= \(\frac {117}{125}\)</p> <p>∴ cos 2\(\alpha\) = \(\frac 7{25}\) and sin 3\(\alpha\) = \(\frac {117}{125}\) <sub>Ans</sub></p>
Q26:
Prove that:
tan(\(\frac {\pi^c}{4}\) - \(\frac A2\)) = \(\sqrt {\frac {1 - sinA}{1 + sinA}}\)
Type: Long
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=tan(\(\frac {\pi^c}{4}\) - \(\frac A2\))</p> <p>= \(\frac {tan\frac \pi4 - tan\frac A2}{1 + tan\frac \pi4 . tan\frac A2}\)</p> <p>= \(\frac {1 - tan\frac A2}{1 + tan\frac A2}\)</p> <p>= \(\frac {1 - \frac {sin\frac A2}{cos\frac A2}}{1 + \frac {sin\frac A2}{cos\frac A2}}\)</p> <p>= \(\frac {\frac {cos\frac A2 - sin\frac A2}{cos\frac A2}}{\frac {cos\frac A2 + sin\frac A2}{cos\frac A2}}\)</p> <p>= \(\frac {cos\frac A2 - sin\frac A2}{cos\frac A2 + sin\frac A2}\)×\(\frac {cos\frac A2 - sin\frac A2}{cos\frac A2 -sin\frac A2}\)</p> <p>= \(\frac {(cos\frac A2 - sin\frac A2)^2}{cos^2\frac A2 - sin^2\frac A2}\)</p> <p>= \(\frac {cos^2\frac A2 + sin^2\frac A2 - 2 sin\frac A2.cos\frac A2}{cosA}\)</p> <p>= \(\frac {1 - sinA}{\sqrt {1 - sin^2A}}\)</p> <p>= \(\sqrt {\frac {(1 - sinA) (1 - sinA)}{(1 + sinA) (1 - sinA)}}\)</p> <p>= \(\sqrt {\frac {1 - sinA}{1 + sinA}}\)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p> <p></p>
Q27:
Prove that:
cot22\(\frac 12\)° - tan22\(\frac 12\)° = 2
Type: Long
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=cot22\(\frac 12\)° - tan22\(\frac 12\)°</p> <p>= \(\frac {cos 22\frac 12°}{sin 22\frac 12°}\) -\(\frac {sin 22\frac 12°}{cos 22\frac 12°}\)</p> <p>= \(\frac {cos^2 22\frac 12° - sin^2 22\frac 12°}{sin 22\frac 12° cos 22\frac 12°}\)</p> <p>= \(\frac {cos 2.22\frac 12°}{\frac 12 × 2 sin 22\frac 12° cos 22\frac 12°}\)</p> <p>= \(\frac {cos 45°}{\frac 12 sin 2.22\frac 12°}\)</p> <p>= \(\frac {2 × \frac 1{\sqrt 2}}{sin 45°}\)</p> <p>= \(\frac {2 × \frac 1{\sqrt 2}}{\frac 1{\sqrt 2}}\)</p> <p>= 2</p> <p>Hence, L.H.S = R.H.S. <sub>Proved</sub></p> <p></p>
Q28:
Find the value of sin 18°.
Type: Long
Difficulty: Easy
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Answer: <p>Let: A = 18°</p> <p>Then,</p> <p>5A = 90°</p> <p>or, 2A + 3A = 90°</p> <p>or, 3A = 90° - 2A</p> <p>Puting cos on both sides:</p> <p>cos 3A = cos (90° - 2A)</p> <p>or, 4 cos<sup>3</sup>A - 3 cosA = sin 2A</p> <p>or, cosA(4 cos<sup>2</sup>A - 3) = 2 sinA cosA</p> <p>or, 4 (1 - sin<sup>2</sup>A) - 3 = 2 sinA</p> <p>or, 4 - 4 sin<sup>2</sup>A - 3 - 2 sinA = 0</p> <p>or, - 4 sin<sup>2</sup>A - 2 sinA + 1 = 0</p> <p>or, -(4 sin<sup>2</sup>A + 2 sinA - 1) = 0</p> <p>or,4 sin<sup>2</sup>A + 2 sinA - 1 = 0</p> <p>Comparing the above condition with ax<sup>2</sup> + bx + c = 0</p> <p>a = 4</p> <p>b = 2</p> <p>c = -1</p> <p>x = \(\frac {-b ± \sqrt {b^2 - 4ac}}{2a}\)</p> <p>Now,</p> <p>sin 18°</p> <p>=\(\frac {-2 ± \sqrt {2^2 - 4 × 4 × (-1)}}{2 × 4}\)</p> <p>=\(\frac {-2 ± \sqrt {4 + 16}}{8}\)</p> <p>=\(\frac {-2 ± \sqrt {20}}{8}\)</p> <p>= \(\frac {2 (-1 + \sqrt 5)}{8}\)</p> <p>= \(\frac {-1 + \sqrt 5}{4}\) <sub>Ans</sub></p> <p></p>
Q29:
Prove that:
tan(\(\frac {\pi}4 + \frac A2\)) = secA + tanA
Type: Long
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=tan(\(\frac {\pi}4 + \frac A2\))</p> <p>= \(\frac {tan \frac \pi4 + tan\frac A2}{1 - tan \frac \pi4 tan \frac A2}\)</p> <p>= \(\frac {tan 45° + tan\frac A2}{1 - tan 45° tan\frac A2}\)</p> <p>= \(\frac {1 + \frac {sin\frac A2}{cos\frac A2}}{1 - 1 × \frac {sin \frac A2}{cos\frac A2}}\)</p> <p>= \(\frac {\frac {cos\frac A2 + sin\frac A2}{cos\frac A2}}{\frac {cos\frac A2 - sin\frac A2}{cos\frac A2}}\)</p> <p>= \(\frac {cos\frac A2 + sin\frac A2}{cos\frac A2 - sin\frac A2}\)× \(\frac {cos\frac A2 + sin\frac A2}{cos\frac A2 + sin\frac A2}\)</p> <p>= \(\frac {(cos\frac A2 + sin\frac A2)^2}{cos^2\frac A2 - sin^2\frac A2}\)</p> <p>= \(\frac {cos^2\frac A2 + sin^2\frac A2 + 2 sin\frac A2 cos\frac A2}{cosA}\)</p> <p>= \(\frac {1 + sinA}{cosA}\)</p> <p>= \(\frac 1{cosA}\) + \(\frac {sinA}{cosA}\)</p> <p>= secA + tanA</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p> <p></p>
Q30:
Prove that:
\(\frac {sin 2A}{1 + cos 2A} × \frac {cosA}{1 + cosA}\) = tan\(\frac A2\)
Type: Long
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=\(\frac {sin 2A}{1 + cos 2A} × \frac {cosA}{1 + cosA}\)</p> <p>= \(\frac {2 sinA cosA}{1 + 2 cos^2A - 1}× \frac {cosA}{1 + cosA}\)</p> <p>= \(\frac {2 sinA cosA}{2 cos^2A}× \frac {cosA}{1 + cosA}\)</p> <p>= \(\frac {sinA}{1 + cosA}\)</p> <p>= \(\frac {2 sin\frac A2 cos\frac A2}{1 + 2 cos^2\frac A2 - 1}\)</p> <p>= \(\frac {2 sin\frac A2 cos\frac A2}{2 cos^2\frac A2}\)</p> <p>= \(\frac {sin\frac A2}{cos\frac A2}\)</p> <p>= tan\(\frac A2\)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q31:
Prove that:
sec(\(\frac \pi4\) + \(\frac \theta2\)) × sec(\(\frac \pi4\) - \(\frac \theta2\)) = 2 sec\(\theta\)
Type: Long
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=sec(\(\frac \pi4\) + \(\frac \theta2\))×sec(\(\frac \pi4\) - \(\frac \theta2\))</p> <p>= \(\frac 1{cos (\frac \pi4 + \frac \theta2)}\)×\(\frac 1{cos (\frac \pi4 - \frac \theta2)}\)</p> <p>= \(\frac 1{(cos\frac \pi4 cos\frac \theta2 - sin\frac \pi4 sin\frac \theta2)(cos\frac \pi4 cos\frac \theta2 + sin\frac \pi4 sin\frac \theta2)}\)</p> <p>= \(\frac 1{(\frac 1{\sqrt 2} cos\frac \theta2 - \frac 1{\sqrt 2} sin\frac \theta2)(\frac 1{\sqrt 2} cos\frac \theta2 + \frac 1{\sqrt 2} sin\frac \theta2)}\)</p> <p>= \(\frac 1{\frac 1{\sqrt 2} (cos\frac \theta2 - sin\frac \theta2) × \frac 1{\sqrt 2} (cos\frac \theta2 + sin\frac \theta2)}\)</p> <p>=\(\frac 1{\frac 12 (cos^2\frac \theta2 - sin^2\frac \theta2)}\)</p> <p>= \(\frac 2{cos\theta}\)</p> <p>= 2 sec\(\theta\)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p> <p></p>
Q32:
Prove that:
cot(\(\frac A2\) + 45°) - tan(\(\frac A2\) + 45°) = \(\frac {2 cosA}{1 + sinA}\)
Type: Long
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=cot(\(\frac A2\) + 45°) - tan(\(\frac A2\) + 45°)</p> <p>= \(\frac {cot\frac A2 cot 45° - 1}{cot 45° + cot\frac A2}\) -\(\frac {tan\frac A2 tan 45°}{1 + tan\frac A2 tan 45°}\)</p> <p>= \(\frac {\frac {cos\frac A2}{sin\frac A2} ×1 - 1}{1 + \frac {cos\frac A2}{sin\frac A2}}\) -\(\frac {\frac {sin\frac A2}{cos\frac A2} - 1}{1 + \frac {sin\frac A2}{cos\frac A2}×1}\)</p> <p>= \(\frac {\frac {cos\frac A2 - sin\frac A2}{sin\frac A2}}{\frac {sin\frac A2 + cos\frac A2}{sin\frac A2}}\) -\(\frac {\frac {sin\frac A2 - cos\frac A2}{cos\frac A2}}{\frac {cos\frac A2 + sin\frac A2}{cos\frac A2}}\)</p> <p>=\(\frac {cos\frac A2 - sin\frac A2}{sin\frac A2 + cos\frac A2}\)- \(\frac {sin\frac A2 - cos\frac A2}{cos\frac A2 + sin\frac A2}\)</p> <p>= \(\frac {cos\frac A2 - sin\frac A2 - sin\frac A2 + cos\frac A2}{cos\frac A2 + sin\frac A2}\)</p> <p>= \(\frac {2(cos\frac A2 - sin\frac A2)}{cos\frac A2 + sin\frac A2}\)× \(\frac {cos\frac A2 - sin\frac A2}{cos\frac A2 - sin\frac A2}\)</p> <p>= \(\frac {2(cos^2\frac A2 + sin^2\frac A2 - 2sin\frac A2 cos\frac A2)}{cos^2\frac A2 - sin^2\frac A2}\)</p> <p>= \(\frac {2(1 - sinA)}{cosA}\)× \(\frac {cosA}{cosA}\)</p> <p>= \(\frac {2(1 - sinA) cosA}{1 - sin^2A}\)</p> <p>= \(\frac {2 cosA (1 - sinA)}{(1 + sinA) (1 - sinA)}\)</p> <p>= \(\frac {2 cosA}{1 + sinA}\)</p> <p>Hence, L.H.S. = R.H.S. <sub>proved</sub></p>
Q33:
Prove that:
tan (\(\frac \pi4\) + \(\frac A2\)) = \(\sqrt {\frac {1 + sinA}{1 - sinA}}\) = \(\frac {1 + sinA}{cosA}\)
Type: Long
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=tan (\(\frac \pi4\) + \(\frac A2\))</p> <p>= \(\frac {tan\frac \pi4 + tan\frac A2}{1 - tan\frac \pi4 tan\frac A2}\)</p> <p>= \(\frac {1 + \frac {sin\frac A2}{cos\frac A2}}{1 -\frac {sin\frac A2}{cos\frac A2}}\)</p> <p>= \(\frac {\frac {cos\frac A2 + sin\frac A2}{cos\frac A2}}{\frac {cos\frac A2 - sin\frac A2}{cos\frac A2}}\)</p> <p>= \(\frac {cos\frac A2 + sin\frac A2}{cos\frac A2 - sin\frac A2}\)</p> <p>= \(\sqrt {\frac {(cos\frac A2 + sin\frac A2)^2}{(cos\frac A2 - sin\frac A2)^2}}\)</p> <p>= \(\sqrt {\frac {cos^2\frac A2 + 2 cos\frac A2 sin\frac A2 + sin^2\frac A2}{cos^2\frac A2 - 2 cos\frac A2 sin\frac A2 + sin^2\frac A2}}\)</p> <p>= \(\sqrt {\frac {1 + sinA}{1 - sinA}}\) M.H.S</p> <p>Again,</p> <p>\(\sqrt {\frac {1 + sinA}{1 - sinA}}\)</p> <p>=\(\sqrt {\frac {1 + sinA}{1 - sinA} × \frac {1 + sinA}{1 + sinA}}\)</p> <p>= \(\sqrt {\frac {(1 + sinA)^2}{cos^2A}}\)</p> <p>= \(\frac {1 + sinA}{cosA}\)</p> <p>Hence, L.H.S = M.H.S = R.H.S. <sub>Proved</sub></p>
Q34:
Prove that:
cos4\(\frac \pi8\) + cos4\(\frac {3\pi}8\) + cos4\(\frac {5\pi}8\) + cos4\(\frac {7\pi}8\) = \(\frac 32\)
Type: Long
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=cos<sup>4</sup>\(\frac \pi8\) + cos<sup>4</sup>\(\frac {3\pi}8\) + cos<sup>4</sup>\(\frac {5\pi}8\) + cos<sup>4</sup>\(\frac {7\pi}8\)</p> <p>=cos<sup>4</sup>\(\frac \pi8\) + cos<sup>4</sup>\(\frac {3\pi}8\) + cos<sup>4</sup>(\(\pi\) - \(\frac {3\pi}8\)) + cos<sup>4</sup>(\(\pi\) - \(\frac {\pi}8\))</p> <p>=cos<sup>4</sup>\(\frac \pi8\) + cos<sup>4</sup>\(\frac {3\pi}8\) + cos<sup>4</sup>\(\frac {3\pi}8\) + cos<sup>4</sup>\(\frac {\pi}8\)</p> <p>= 2 cos<sup>4</sup>\(\frac \pi8\) + 2 cos<sup>4</sup>\(\frac {3\pi}8\)</p> <p>= 2 (cos<sup>4</sup>\(\frac \pi8\) + cos<sup>4</sup>\(\frac {3\pi}8\))</p> <p>= 2 [cos<sup>4</sup>\(\frac \pi8\) + cos<sup>4</sup>(\(\frac{\pi}2 - \frac {\pi}8\))]</p> <p>= 2 [cos<sup>4</sup>\(\frac \pi8\) + sin<sup>4</sup>\(\frac \pi8\)]</p> <p>= 2 [(cos<sup>2</sup>\(\frac \pi8\))<sup>2</sup> + (sin<sup>2</sup>\(\frac \pi8\))<sup>2</sup>]</p> <p>= 2 [(cos<sup>2</sup>\(\frac \pi8\) + sin<sup>2</sup>\(\frac \pi8\))<sup>2</sup> - 2cos<sup>2</sup>\(\frac \pi8\)sin<sup>2</sup>\(\frac \pi8\)]</p> <p>= [2 - 4 cos<sup>2</sup>\(\frac \pi8\)sin<sup>2</sup>\(\frac \pi8\)]</p> <p>= 2 - (2 sin\(\frac \pi8\)cos\(\frac \pi8\) )<sup>2</sup></p> <p>= 2 - (sin 2 . \(\frac \pi8\))<sup>2</sup></p> <p>= 2 - (sin \(\frac \pi4\))<sup>2</sup></p> <p>= 2 - (\(\frac 1{\sqrt 2}\))<sup>2</sup></p> <p>= 2 - \(\frac 12\)</p> <p>= \(\frac {4 - 1}2\)</p> <p>= \(\frac 32\)</p> <p>Hence, L.H.S. = R.H.S <sub>Proved</sub></p> <p></p>
Q35:
Without using table or calculator. Find the value of sin 18° and cos 36°.
Type: Long
Difficulty: Easy
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Answer: <p>Let:</p> <p>\(\theta\) = 18°</p> <p>Then:</p> <p>5\(\theta\) = 90°</p> <p>i.e. 2\(\theta\) + 3\(\theta\) = 5\(\theta\)</p> <p>or, 2\(\theta\) + 3\(\theta\) = 90°</p> <p>or, 2\(\theta\) = 90° - 3\(\theta\)</p> <p>Now,</p> <p>sin 2\(\theta\) = sin (90° - 3\(\theta\))</p> <p>or, 2 sin\(\theta\) cos\(\theta\) = 4 cos<sup>3</sup>\(\theta\) - 3 cos\(\theta\)</p> <p>or,2 sin\(\theta\) cos\(\theta\) = cos\(\theta\) (4 cos<sup>2</sup>\(\theta\) - 3)</p> <p>or, 2 sin\(\theta\) = 4 - 4 sin<sup>2</sup>\(\theta\) - 3</p> <p>or, 2 sin\(\theta\) = 1 - 4sin<sup>2</sup>\(\theta\)</p> <p>or, 4sin<sup>2</sup>\(\theta\) +2 sin\(\theta\) - 1 = 0 .................................(i)</p> <p>Comparingequation (i) with quadratic equation ax<sup>2</sup>+ bx + c = 0, we get:</p> <p>sin\(\theta\)</p> <p>= \(\frac {-b ± \sqrt {b^2 - 4ac}}{2a}\)</p> <p>= \(\frac {-2 ± \sqrt {2^2 - 4 . 4 . (-1)}}{2 . 4}\)</p> <p>= \(\frac {-2 ± \sqrt {4 + 16}}{8}\)</p> <p>= \(\frac {-2 ± \sqrt {20}}{8}\)</p> <p>= \(\frac {-2 ± 2\sqrt 5}{8}\)</p> <p>= \(\frac {-1 ± \sqrt 5}{4}\)</p> <p>∴ sin 18° =\(\frac {-1 ± \sqrt 5}{4}\)</p> <p>But the value of sin 18° is positive.</p> <p>Put (+) sign</p> <p>∴ sin 18° =\(\frac {-1 + \sqrt 5}{4}\)</p> <p>And</p> <p>cos 36°</p> <p>= 1 - 2 sin<sup>2</sup>\(\frac {36°}2\) [\(\because\) cos\(\theta\) = 1 - sin<sup>2</sup>\(\frac \theta2\)]</p> <p>= 1 - 2 sin<sup>2</sup>18°</p> <p>= 1 - 2 (\(\frac {-1 + \sqrt 5}{4}\))<sup>2</sup></p> <p>= 1 - 2 (\(\frac {1 - 2\sqrt 5 + 5}{16}\))</p> <p>= 1 - (\(\frac {6 - 2\sqrt 5}8\))</p> <p>= \(\frac {8 - 6 + 2\sqrt 5}{8}\)</p> <p>= \(\frac {2 + 2\sqrt 5}{8}\)</p> <p>= \(\frac {2 (1 + \sqrt 5)}8\)</p> <p>= \(\frac {1 + \sqrt 5}4\)</p> <p>∴ sin 18° =\(\frac {-1 + \sqrt 5}{4}\)</p> <p>and cos 36° =\(\frac {1 + \sqrt 5}{4}\) <sub>Ans</sub></p> <p></p>
