Subjective Questions
Q1:
Prove that:
sin 50° - sin 70° + sin 10° = 0
Type: Short
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=sin 50° - sin 70° + sin 10°</p> <p>= 2 cos(\(\frac {50° + 70°}2\)) sin (\(\frac {50° - 70°}2\)) + sin 10°</p> <p>= 2 cos(\(\frac {120°}2\)) sin(\(\frac {-20°}2\)) + sin 10°</p> <p>= - 2 cos 60° sin 10° + sin 10°</p> <p>= - 2× \(\frac 12\) sin 10° + sin 10°</p> <p>= - sin 10° + sin 10°</p> <p>= 0</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q2:
Express sin 36° sin 24° as difference.
Type: Short
Difficulty: Easy
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Answer: <p>sin 36° sin 24°</p> <p>= \(\frac 12\) [2 sin 36° sin 24°]</p> <p>= \(\frac 12\) [cos (36° - 24°) - cos(36° + 24°)]</p> <p>= \(\frac 12\) [cos 12° - cos 60°] <sub>Ans</sub></p>
Q3:
Express sin 50° cos 32° as sum.
Type: Short
Difficulty: Easy
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Answer: <p>sin 50° cos 32°</p> <p>= \(\frac 12\) (2 sin 50° cos 32°)</p> <p>=\(\frac 12\) [sin (50° + 32°) + sin (50° - 32°)]</p> <p>= \(\frac 12\)[sin 82° + sin 18°] <sub>Ans</sub></p>
Q4:
Prove that:
\(\frac {cosB - cosA}{cosA + cosB}\) = tan\(\frac {A + B}{2}\) tan\(\frac {A - B}2\)
Type: Short
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=\(\frac {cosB - cosA}{cosA + cosB}\)</p> <p>= \(\frac {2 sin(\frac {B + A}2) sin(\frac {A - B}2)}{2 cos(\frac {A + B}2) cos(\frac {A - B}2)}\)</p> <p>= tan\(\frac {A + B}{2}\) tan\(\frac {A - B}2\)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q5:
Prove that:
\(\frac {sinA + sinB}{cosA + cosB}\) = tan(\(\frac {A + B}2\))
Type: Short
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=\(\frac {sinA + sinB}{cosA + cosB}\)</p> <p>= \(\frac {2 sin(\frac {A + B}2) cos(\frac {A - B}2)}{2 cos(\frac {A + B}2) cos(\frac {A - B}2)}\)</p> <p>= \(\frac {sin(\frac {A + B}2)}{cos(\frac {A + B}2)}\)</p> <p>= tan(\(\frac {A + B}2\))</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p> <p></p>
Q6:
Prove that:
\(\frac {sin 5A - sin 3A}{cos 5A + cos 3A}\) = tanA
Type: Short
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=\(\frac {sin 5A - sin 3A}{cos 5A + cos 3A}\)</p> <p>=\(\frac {2 cos(\frac {5A + 3A}2) sin(\frac {5A - 3A}2)}{2 cos(\frac {5A + 3A}2) cos(\frac {5A - 3A}2)}\)</p> <p>= \(\frac {cos4A sinA}{cos 4A cosA}\)</p> <p>= \(\frac {sinA}{cosA}\)</p> <p>= tanA</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p> <p></p>
Q7:
Evaluate without using calculator or table.
sin 70° - cos 80° + cos 140°
Type: Short
Difficulty: Easy
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Answer: <p>sin 70° - cos 80° + cos 140°</p> <p>= sin 70° + cos 140° - cos 80°</p> <p>= sin 70° - 2 sin\(\frac {140° + 80°}2\) sin\(\frac {140° - 80°}2\)</p> <p>= sin 70° - 2 sin\(\frac {220°}2\) sin\(\frac {60°}2\)</p> <p>= sin 70° - 2 sin 110° sin 30°</p> <p>= sin 70° - 2 sin(180° - 70°)× \(\frac 12\)</p> <p>= sin 70° - sin 70°</p> <p>= 0 <sub>Ans</sub></p>
Q8:
Without using calculator or table, prove that:
cos 70° + cos 40° = 2 cos 55° cos 15°
Type: Short
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>= cos 70° + cos 40°</p> <p>= 2 cos\(\frac {70° + 40°}2\) cos\(\frac {70° - 40°}2\)</p> <p>= 2 cos\(\frac {110°}2\) cos\(\frac {30°}2\)</p> <p>= 2 cos 55° cos 15°</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q9:
Without using calculator or table, find the value of cos 15° - cos 75°.
Type: Short
Difficulty: Easy
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Answer: <p>Here,</p> <p>cos 15° - cos 75°</p> <p>= - 2 sin\(\frac {15 + 75}2\) sin\(\frac {15 - 75}2\)</p> <p>= - 2 sin\(\frac {90°}2\) sin\(\frac {(-60°)}2\)</p> <p>= - 2 sin 45°× - sin 30°</p> <p>= 2× \(\frac 1{\sqrt 2}\)× \(\frac 12\)</p> <p>= \(\frac 1{\sqrt 2}\) <sub>Ans</sub></p>
Q10:
Express as a product sin 50° + sin 20°.
Type: Short
Difficulty: Easy
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Answer: <p>Here,</p> <p>sin 50° + sin 20°</p> <p>= 2 sin\(\frac {50° + 20°}2\) cos\(\frac {50° - 20°}2\)</p> <p>= 2 sin\(\frac {70°}2\) cos\(\frac {30°}2\)</p> <p>= 2 sin 35° cos 15° <sub>Ans</sub></p>
Q11:
Express as a sum or difference: sin 25° cos 75°.
Type: Short
Difficulty: Easy
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Answer: <p>Here,</p> <p>sin 25° cos 75°</p> <p>= \(\frac 22\) sin 25° cos 75°</p> <p>= \(\frac 12\) (2 sin 25° cos 75°)</p> <p>= \(\frac 12\) [sin (25° + 75°) + sin (25° - 75°)]</p> <p>= \(\frac 12\) [sin 100° + sin (-50°)]</p> <p>= \(\frac 12\) [sin 100° - sin 50°] <sub>Ans</sub></p>
Q12:
Prove that:
sin 5\(\theta\) + sin 3\(\theta\) = 2 sin 4\(\theta\) cos\(\theta\)
Type: Short
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=sin 5\(\theta\) + sin 3\(\theta\)</p> <p>= 2 sin\(\frac {5\theta + 3\theta}2\) cos\(\frac {5\theta - 3\theta}2\)</p> <p>= 2 sin\(\frac {8\theta}2\) cos\(\frac {2\theta}2\)</p> <p>= 2 sin 4\(\theta\) cos\(\theta\)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p> <p></p>
Q13:
Prove that:
\(\frac {sinA + sinB}{sinA - sinB}\) = tan\(\frac {A + B}2\) cot\(\frac {A - B}2\)
Type: Short
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=\(\frac {sinA + sinB}{sinA - sinB}\)</p> <p>= \(\frac {2 sin(\frac {A + B}2) cos(\frac {A - B}2)}{2 cos(\frac {A + B}2) sin(\frac {A - B}2)}\)</p> <p>= tan\(\frac {A + B}2\) cot\(\frac {A - B}2\)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q14:
Prove that:
\(\frac {sin 3A - sinA}{cosA - cos 3A}\) = cot 2A
Type: Short
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=\(\frac {sin 3A - sinA}{cosA - cos 3A}\)</p> <p>= \(\frac {2 cos\frac {3A + A}2 . sin\frac {3A - A}2}{2 sin\frac {3A + A}2 . sin\frac {3A - A}2}\)</p> <p>= \(\frac {cos\frac {4A}2 . sin\frac {2A}2}{sin\frac {4A}2 . sin\frac {2A}2}\)</p> <p>= \(\frac {cos 2A}{sin 2A}\)</p> <p>= cot 2A</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p> <p></p>
Q15:
Prove that:
\(\frac {sin 2A + sin 5A - sinA}{cos 2A + cos 5A + cosA}\) = tan 2A
Type: Short
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=\(\frac {sin 2A + sin 5A - sinA}{cos 2A + cos 5A + cosA}\)</p> <p>= \(\frac {sin 2A + 2 cos\frac {5A + A}2 . sin\frac {5A - A}2}{cos 2A + 2 cos\frac {5A + A}2 . cos\frac {5A - A}2}\)</p> <p>= \(\frac {sin 2A + 2 cos 3A . sin 2A}{cos 2A + 2 cos3A . cos 2A}\)</p> <p>= \(\frac {sin 2A (1 + 2 cos 3A)}{cos 2A (1 + 2 cos 3A)}\)</p> <p>= \(\frac {sin 2A}{cos 2A}\)</p> <p>= tan 2A</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p> <p></p>
Q16:
Prove that:
sin 36° sin 72° sin 108° sin 144° = \(\frac 5{16}\)
Type: Long
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=sin 36° sin 72° sin 108° sin 144°</p> <p>= \(\frac 12\)(2 sin 36° sin 144°)× \(\frac 12\)(2 sin 72° sin 108°)</p> <p>= \(\frac 14\) [cos(36° - 144°) - cos(36° + 144°)] [cos (72° - 108°) - cos (72° + 108°)]</p> <p>= \(\frac 14\) [cos(-108°) - cos(180°)] [cos(-36°) - cos(108°)]</p> <p>= \(\frac 14\) [cos (90° + 18°) - (-1)] [cos 36° - (-1)] [\(\because\) cos (-\(\theta\)) = cos\(\theta\)]</p> <p>= \(\frac 14\) [sin 18° + 1] [cos 36° + 1]</p> <p>= \(\frac 14\) [-\(\frac {\sqrt 5 - 1}4\) + 1] [\(\frac {\sqrt 5 + 1}4\) + 1]</p> <p>[\(\because\) sin 18° = \(\frac {-\sqrt 5 - 1}4\), cos 36° = \(\frac {\sqrt 5 + 1}4\)]</p> <p>= \(\frac 14\) (\(\frac {-\sqrt 5 + 1 + 4}4\)) (\(\frac {\sqrt 5 + 1 + 4}4\))</p> <p>= \(\frac 14\) (\(\frac {5 - \sqrt 5}4\)) (\(\frac {5 + \sqrt 5}4\))</p> <p>= \(\frac 14\)× \(\frac {(5)^2 - (\sqrt 5)^2}{16}\)</p> <p>= \(\frac 14\)× \(\frac {25 - 5}{16}\)</p> <p>= \(\frac {20}{4 × 16}\)</p> <p>= \(\frac 56\)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q17:
Prove that:
cosA cos(60° - A) cos(60° + A) = \(\frac 14\) cos 3A
Type: Long
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=cosAcos(60° - A) cos(60° + A)</p> <p>= \(\frac 12\) cosA [2cos(60° - A) cos(60° + A)]</p> <p>= \(\frac 12\) cosA [cos(60° + A + 60° - A) + cos(60° + A - 60° + A )]</p> <p>= \(\frac 12\) cosA [cos 120° + co 2A]</p> <p>= \(\frac 12\) cosA× (-\(\frac 12\)) + \(\frac 12\) cosA cos 2A</p> <p>= -\(\frac 14\) cosA + \(\frac 12\)× \(\frac 12\) (2 cosA cos 2A)</p> <p>= -\(\frac 14\) cosA + \(\frac 14\) [cos(2A + A) + cos(2A - A)]</p> <p>= -\(\frac 14\) cosA + \(\frac 14\) (cos 3A + cosA)</p> <p>= -\(\frac 14\) cosA + \(\frac 14\) cos 3A + \(\frac 14\) cosA</p> <p>= \(\frac 14\) cos 3A</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q18:
Find the value of sin 20° sin 40° sin 80°.
Type: Long
Difficulty: Easy
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Answer: <p>Here,</p> <p>sin 20° sin 40° sin 80°</p> <p>= sin 20° . \(\frac 12\) [2 sin 40° sin 80°]</p> <p>= \(\frac 12\) sin 20° [cos(40° - 80°) - cos(40° + 80°)]</p> <p>= \(\frac 12\) sin 20° [cos (-40°) - cos 120°]</p> <p>= \(\frac 12\) sin 20° cos 40° + \(\frac 14\) sin 20° [\(\because\) cos(-\(\theta\)) = cos\(\theta\)]</p> <p>= \(\frac 12\)× \(\frac 12\) [2 sin 20° cos 40°] + \(\frac 14\)sin 20°</p> <p>= \(\frac 14\) [sin (20° + 40°) + sin (20° - 40°)] + \(\frac 14\)sin 20°</p> <p>= \(\frac 14\) [sin 60° + sin (-20°)] + \(\frac 14\)sin 20°</p> <p>= \(\frac 14\) (\(\frac {\sqrt 3}2\) - sin 20°) + \(\frac 14\) sin 20° [\(\because\) sin(-\(\theta\)) = - sin\(\theta\)]</p> <p>= \(\frac {\sqrt 3}8\) - \(\frac 14\)sin 20° + \(\frac 14\)sin 20°</p> <p>= \(\frac {\sqrt 3}8\)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q19:
Without using calculator or table, find the numerical value of:
8 sin 20° . sin 40° . sin 80°
Type: Long
Difficulty: Easy
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Answer: <p>Here,</p> <p>8 sin 20° . sin 40° . sin 80°</p> <p>= 4 sin 20°× (2 sin 40° sin 80°)</p> <p>= 4 sin 20° [cos (40° - 80°) - cos (40° + 80°)]</p> <p>= 4 sin 20° [cos (-40°) - cos 120°]</p> <p>= 4 sin 20° [cos 40° - (-\(\frac 12\))] [\(\because\) cos (-\(\theta\)) = cos\(\theta\)]</p> <p>= 4 sin 20° cos 40° + 2 sin 20°</p> <p>= 2× (2 sin 20° cos 40°) + 2 sin 20°</p> <p>= 2 [sin (20° + 40°) + sin (20° - 40°)] + 2 sin 20°</p> <p>= 2 [sin 60° + sin (-20°)] + 2 sin 20°</p> <p>= 2 [\(\frac {\sqrt 3}2\) - sin 20°] + 2 sin 20° [\(\because\) sin (-\(\theta\)) = - sin\(\theta\)]</p> <p>= \(\sqrt 3\) - 2 sin 20° + 2 sin 20°</p> <p>= \(\sqrt 3\) <sub>Ans</sub></p>
Q20:
Prove that:
cos 20° cos 40° cos 60° cos 80° = \(\frac 1{16}\)
Type: Long
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>= cos 20° cos 40° cos 60° cos 80°</p> <p>= cos 20° cos 40° \(\frac 12\) cos 80°</p> <p>= \(\frac 14\) cos 20° (cos 40° cos 80°)</p> <p>= \(\frac 14\) cos 20° [cos (40° + 80°) + cos (40° - 80°)]</p> <p>= \(\frac 14\) cos 20° [cos 120° + cos (-40°)]</p> <p>= \(\frac 14\) cos 20° [cos 120° + cos 40°]</p> <p>= \(\frac 14\) cos 20° [-\(\frac 12\) + cos 40°]</p> <p>= -\(\frac 18\) cos 20° + \(\frac 14\) cos 20° cos 40°</p> <p>= -\(\frac 18\) cos 20° + \(\frac 18\) [2 cos 20° cos 40°]</p> <p>= -\(\frac 18\) cos 20° + \(\frac 18\) [cos (20° + 40°) + cos (20° - 40°)]</p> <p>= -\(\frac 18\) cos 20° + \(\frac 18\) [cos 60° + cos (-20°)]</p> <p>= -\(\frac 18\) cos 20° + \(\frac 18\) [\(\frac 12\) + cos 20°]</p> <p>= -\(\frac 18\) cos 20° + \(\frac 1{16}\) + \(\frac 18\) cos 20°</p> <p>= \(\frac 1{16}\)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q21:
Prove that:
cos 20° cos 30° cos 40° cos 80° = \(\frac {\sqrt 3}{16}\)
Type: Long
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>= cos 20° cos 30° cos 40° cos 80°</p> <p>= cos 30°× \(\frac 12\) cos 20° (2 cos 40° cos 80°)</p> <p>= \(\frac {\sqrt 3}{2}\) cos 20° [cos 120° + cos (-40°)]</p> <p>= \(\frac {\sqrt 3}4\) cos 20° [(-\(\frac 12\)) + cos 40°]</p> <p>= -\(\frac {\sqrt 3}8\) cos 20° + \(\frac {\sqrt 3}4\) cos 20° cos 40°</p> <p>= -\(\frac {\sqrt 3}8\) cos 20° + \(\frac {\sqrt 3}4\)× \(\frac 12\) (2 cos 20° cos 40°)</p> <p>= -\(\frac {\sqrt 3}8\) cos 20° + \(\frac {\sqrt 3}8\) cos (20° + 40°) + cos (20° - 40°)</p> <p>= -\(\frac {\sqrt 3}8\) cos 20° + \(\frac {\sqrt 3}8\) (cos 60° + cos (-20°))</p> <p>= -\(\frac {\sqrt 3}8\) cos 20° + \(\frac {\sqrt 3}{16}\) + \(\frac {\sqrt 3}8\) cos 20°</p> <p>= \(\frac {\sqrt 3}{16}\)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q22:
Prove that:
cos2\(\theta\) + cos2(\(\theta\) - 120°) + cos2(\(\theta\) + 120°) = \(\frac 32\)
Type: Long
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=cos<sup>2</sup>\(\theta\) + cos<sup>2</sup>(\(\theta\) - 120°) + cos<sup>2</sup>(\(\theta\) + 120°)</p> <p>= \(\frac 12\) [2 cos<sup>2</sup>\(\theta\) + 2 cos<sup>2</sup>(\(\theta\) + 120°) + 2 cos<sup>2</sup>(\(\theta\) + 120°)]</p> <p>= \(\frac 12\) [1 + cos 2\(\theta\) + 1 + cos 2(\(\theta\) - 120°) + 1 + cos 2(\(\theta\) + 120°)]</p> <p>= \(\frac 12\) [3 + cos 2\(\theta\) + cos (2\(\theta\) + 240°) + cos (2\(\theta\) + 240°)]</p> <p>= \(\frac 12\) [3 + cos 2\(\theta\) + 2 cos (\(\frac {2\theta - 240° + 2\theta + 240°}2\)) cos (\(\frac {2\theta - 240° - 2\theta - 240°}2\))]</p> <p>= \(\frac 12\) [3 + cos 2\(\theta\) + 2 cos\(\frac {4\theta}2\) cos\(\frac {-480°}2\)]</p> <p>= \(\frac 12\) [3 + cos 2\(\theta\) + 2 cos 2\(\theta\) cos 240°]</p> <p>= \(\frac 12\) [3 + cos 2\(\theta\) + 2 cos 2\(\theta\) × -\(\frac 12\)]</p> <p>= \(\frac 12\) [3 + cos 2\(\theta\) - cos 2\(\theta\)]</p> <p>= \(\frac 32\)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p> <p></p>
Q23:
Prove that:
sin\(\theta\) sin (60° - \(\theta\)) . sin (60° + \(\theta\)) = \(\frac 14\) sin 3\(\theta\)
Type: Long
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=sin\(\theta\) sin (60° - \(\theta\)) . sin (60° + \(\theta\))</p> <p>= sin\(\theta\) \(\frac 12\) [2 sin (60° - \(\theta\)) sin (60° + \(\theta\))]</p> <p>= sin\(\theta\) \(\frac 12\) [cos (60° - \(\theta\) - 60° - \(\theta\)) - cos (60° - \(\theta\) + 60° + \(\theta\))]</p> <p>= sin\(\theta\) \(\frac 12\) [cos (-2\(\theta\)) - cos 120°]</p> <p>= \(\frac 12\) sin\(\theta\) [cos 2\(\theta\) - (-\(\frac 12\))]</p> <p>= \(\frac 12\) sin\(\theta\) cos 2\(\theta\) + \(\frac 14\) sin\(\theta\)</p> <p>= \(\frac 12\)× \(\frac 12\) [2 sin\(\theta\) cos 2\(\theta\)] + \(\frac 14\) sin\(\theta\)</p> <p>= \(\frac 14\) [sin (\(\theta\) + 20°) - sin (\(\theta\) - 2\(\theta\))] + \(\frac 14\) sin\(\theta\)</p> <p>= \(\frac 14\) [sin 3\(\theta\) + sin(-\(\theta\))] + \(\frac 14\) sin\(\theta\)</p> <p>= \(\frac 14\) sin 3\(\theta\) - \(\frac 14\) sin\(\theta\) + \(\frac 14\)sin\(\theta\)</p> <p>= \(\frac 14\) sin 3\(\theta\)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q24:
Prove that:
(cosA + cosB)2 + (sinA + sinB)2 = 4 cos2\(\frac {A - B}2\)
Type: Long
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=(cosA + cosB)<sup>2</sup> + (sinA + sinB)<sup>2</sup></p> <p>= (2 cos\(\frac {A + B}2\) . cos\(\frac {A - B}2\))<sup>2</sup> +(2 sin\(\frac {A + B}2\) . cos\(\frac {A - B}2\))<sup>2</sup></p> <p>= 4 cos<sup>2</sup>\(\frac {A + B}2\) . cos<sup>2</sup>\(\frac {A - B}2\) +4 sin<sup>2</sup>\(\frac {A + B}2\) . cos<sup>2</sup>\(\frac {A - B}2\)</p> <p>= 4 cos<sup>2</sup>\(\frac {A - B}2\) (cos<sup>2</sup>\(\frac {A + B}2\) + sin<sup>2</sup>\(\frac {A + B}2\))</p> <p>=4 cos<sup>2</sup>\(\frac {A - B}2\)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q25:
If sin\(\alpha\) = k sin\(\beta\), prove that:
\(\frac {\alpha - \beta}2\) = \(\frac {k - 1}{k + 1}\) tan\(\frac {\alpha + \beta}2\)
Type: Long
Difficulty: Easy
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Answer: <p>Given:</p> <p>\(\frac {sin\alpha}{sin\beta}\) = \(\frac k1\)</p> <p>By componendo and dividend, we get:</p> <p>\(\frac {sin\alpha + sin\beta}{sin\alpha - sin\beta}\) = \(\frac {k + 1}{k - 1}\)</p> <p>or, \(\frac {2 sin\frac {\alpha + \beta}2 . cos\frac {\alpha - \beta}2}{2 cos\frac {\alpha + \beta}2 . sin\frac {\alpha - \beta}2}\) =\(\frac {k + 1}{k - 1}\)</p> <p>or, tan\(\frac {\alpha + \beta}2\) . cot\(\frac {\alpha - \beta}2\) =\(\frac {k + 1}{k - 1}\)</p> <p>or, tan\(\frac {\alpha + \beta}2\)=\(\frac {k + 1}{k - 1}\) . \(\frac 1{cot\frac {\alpha - \beta}2}\)</p> <p>or,tan\(\frac {\alpha + \beta}2\) .\(\frac {k + 1}{k - 1}\) = tan\(\frac {\alpha - \beta}2\)</p> <p>∴tan\(\frac {\alpha - \beta}2\) =\(\frac {k + 1}{k - 1}\) .tan\(\frac {\alpha + \beta}2\)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q26:
Prove that:
\(\frac {cos 7\theta + cos 3\theta - cos 5\theta - cos\theta}{sin 7\theta - sin 3\theta - sin 5\theta +sin\theta}\) = cot 2\(\theta\)
Type: Long
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=\(\frac {cos 7\theta + cos 3\theta - cos 5\theta - cos\theta}{sin 7\theta - sin 3\theta - sin 5\theta +sin\theta}\)</p> <p>= \(\frac {(cos 7\theta + cos 3\theta) - (cos 5\theta + cos\theta)}{(sin 7\theta - sin 3\theta) - (sin 5\theta - sin\theta)}\)</p> <p>= \(\frac {2 cos\frac {7\theta + 3\theta}2 . cos\frac {7\theta - 3\theta}2 - 2 cos\frac {5\theta + \theta}2 . cos\frac {5\theta - \theta}2}{2 cos\frac {7\theta + 3\theta}2 . sin\frac {7\theta - 3\theta}2 - 2 cos\frac {5\theta + \theta}2 . sin\frac {5\theta - \theta}2}\)</p> <p>= \(\frac {2 cos 5\theta. cos 2\theta - 2 cos 3\theta. cos 2\theta}{2 cos 5\theta. sin 2\theta - 2 cos 3\theta. sin 2\theta}\)</p> <p>= \(\frac {2 cos 2\theta (cos 5\theta - cos 3\theta)}{2 sin 2\theta (cos 5\theta - cos 3\theta)}\)</p> <p>= cot 2\(\theta\)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q27:
Prove that:
\(\frac {cos 3A + 2 cos 5A + cos 7A}{cosA + 2 cos 3A + cos 5A}\) = cos 2A - sin 2A . tan 3A
Type: Long
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=\(\frac {cos 3A + 2 cos 5A + cos 7A}{cosA + 2 cos 3A + cos 5A}\)</p> <p>= \(\frac {2 cos 5A + (cos 3A + cos 7A)}{2 cos 3A + (cosA + cos 5A)}\)</p> <p>= \(\frac {2 cos 5A + 2 cos\frac {3A + 7A}2 . cos\frac {3A - 7A}2}{2 cos 3A + 2 cos\frac {A + 5A}2 . cos\frac {A - 5A}2}\)</p> <p>= \(\frac {2 cos 5A + 2 cos 5A . cos (-2A)}{cos 3A + 2 cos 3A . cos (-2A)}\)</p> <p>= \(\frac {2 cos 5A + 2 cos 5A . cos 2A}{2 cos 3A + 2 cos 3A . cos 2A}\)</p> <p>= \(\frac {2 cos 5A (1 + cos 2A)}{2 cos 3A (1 + cos 2A)}\)</p> <p>= \(\frac {cos 5A}{cos 3A}\)</p> <p>= \(\frac {cos (2A + 3A)}{cos 3A}\)</p> <p>= \(\frac {cos 2A . cos 3A - sin 2A . sin 3A}{cos 3A}\)</p> <p>= \(\frac {cos 2A . cos 3A}{cos 3A}\) - \(\frac {sin 2A . sin 3A}{cos 3A}\)</p> <p>= cos 2A - sin 2A . tan 3A</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p> <p></p>
Q28:
Prove that:
cos2x . sin4x = \(\frac 1{32}\) (cos 6x - 2 cos 4x - cos 2x + 2)
Type: Long
Difficulty: Easy
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Answer: <p>L.H.S.</p> <p>=cos<sup>2</sup>x . sin<sup>4</sup>x</p> <p>= cos<sup>2</sup>x . sin<sup>2</sup>x . sin<sup>2</sup>x</p> <p>= sin<sup>2</sup>x \(\frac 14\) [4 sin<sup>2</sup>x . cos<sup>2</sup>x]</p> <p>= sin<sup>2</sup>x \(\frac 14\) [(2 sinx . cosx)<sup>2</sup>]</p> <p>= \(\frac 14\) sin<sup>2</sup>x . sin<sup>2</sup>2x</p> <p>= \(\frac 14\) sin<sup>2</sup>x . \(\frac 12\) . 2 sin<sup>2</sup>2x</p> <p>= \(\frac 18\) sin<sup>2</sup>x (2 sin<sup>2</sup>2x)</p> <p>= \(\frac 18\) sin<sup>2</sup>x (1 - cos 4x)</p> <p>= \(\frac 18\) sin<sup>2</sup>x - \(\frac 18\) sin<sup>2</sup>x . cos 4x</p> <p>= \(\frac 18\) sin<sup>2</sup>x - \(\frac 18\) (1 - cos<sup>2</sup>x) . cos 4x</p> <p>= \(\frac 18\) sin<sup>2</sup>x - \(\frac 18\) cos 4x + \(\frac 18\) cos<sup>2</sup>x . cos 4x</p> <p>= \(\frac 18\) sin<sup>2</sup>x - \(\frac 18\) cos 4x + \(\frac 1{8}\) × \(\frac 12\) [2 cos<sup>2</sup>x .cos 4x]</p> <p>= \(\frac 18\) sin<sup>2</sup>x - \(\frac 18\) cos 4x + \(\frac 1{16}\) cos 4x (1 + cos 2x)</p> <p>= \(\frac 18\) sin<sup>2</sup>x - \(\frac 18\) cos 4x + \(\frac 1{16}\) cos 4x + \(\frac 1{16}\) cos 4x . cos 2x</p> <p>= \(\frac 18\) sin<sup>2</sup>x + \(\frac 1{16}\) cos 4x + \(\frac 1{32}\) 2 cos 4x . cos 2x - \(\frac 18\) cos 4x</p> <p>= \(\frac 18\) sin<sup>2</sup>x - \(\frac 1{16}\) cos 4x + \(\frac 1{32}\)[cos (4x + 2x) + cos (4x - 2x)]</p> <p>= \(\frac 18\) sin<sup>2</sup>x - \(\frac 1{16}\) cos 4x + \(\frac 1{32}\)(cos 6x + cos 2x)</p> <p>= \(\frac 1{16}\) 2 sin<sup>2</sup>x - \(\frac 1{16}\) cos 4x + \(\frac 1{32}\) cos 6x + \(\frac 1{32}\) cos 2x</p> <p>= \(\frac 1{16}\) (1 - cos 2x) - \(\frac 1{16}\) cos 4x + \(\frac 1{32}\) cos 6x + \(\frac 1{32}\) cos 2x</p> <p>= \(\frac 1{16}\) - \(\frac 1{16}\) cos 2x -\(\frac 1{16}\) cos 4x + \(\frac 1{32}\) cos 6x + \(\frac 1{32}\) cos 2x</p> <p>= \(\frac 1{32}\) (2 - 2 cos 2x - 2 cos 4x + cos 6x + cos 2x)</p> <p>= \(\frac 1{32}\) (cos 6x - 2 cos 4x - cos 2x + 2)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q29:
a) sin 750 + sin 150
b) sin750 - sin1050
c) cos 150 - cos750
d) sin700 - cos 800 + cos 1400
Type: Short
Difficulty: Easy
Q30:
cos 10\(5^o\) + cos 1\(5^o\) = \(\frac{1}{\sqrt 2}\)
Type: Short
Difficulty: Easy
Q31:
cos7\(0^o\)+cos4\(0^o\)=2 cos 5\(5^o\) . cos1\(5^o\)
Type: Short
Difficulty: Easy
Q32:
$$ \cos 75^0 + \cos 15^0 = \sqrt {\frac{ 3}{2 }}$$
Type: Short
Difficulty: Easy
Q33:
$$ \sin50^o + sin70^o=\sqrt3 cos10^o$$
Type: Short
Difficulty: Easy
Q34:
$$ \cos40^o+sin40^o=\sqrt2cos5^o$$
Type: Short
Difficulty: Easy
Q35:
$$ \cos40^o+sin\:40^o=\sqrt2\:cos\:5^o$$
Type: Short
Difficulty: Easy
Q36:
$$ \sin65^o\:+\:cos\:65^o=\sqrt2\: cos20^o$$
Type: Short
Difficulty: Easy
It is the supplements of main memory. It is mainly used to transfer data or program from one computer to another computer. It also functions as back up devices which allows backing up the valuable information that you are working on i.e. storing data for future purpose. The most common types of auxiliary storage devices are magnetic tapes, magnetic disks, floppy disks, hard disks etc.
