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Evaluate: cos 58^@/sin 32^@ + sin 22^@/cos 68^@ – (cos 38^@ cosec 52^@)/(tan 18^@ tan 35^@ tan 60^@ tan 72^@ tan 55^@)

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Question

Evaluate: `cos 58^@/sin 32^@  + sin 22^@/cos 68^@  - (cos 38^@ cosec 52^@)/(tan 18^@ tan 35^@ tan 60^@ tan 72^@ tan 55^@)`

Evaluate
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Solution

cos 58° = cos (90° – 32°) = sin 32°

sin 22° = sin (90° – 68°) = cos 68°

cos 38° = cos (90 – 52) = sin 52°

tan 18° = cot 72 tan 35° = cot 55°

`=> sin 32^@/sin 32^@ + cos 68^@/cos 68^@ - (sin 52 cosec 52)/(tan 72.cot72 tan 55 cot 55.tan 60)`

`= 1 + 1 - 1/sqrt3 = (2sqrt3-1)/sqrt3 xx sqrt3/sqrt3 = (6 - sqrt3)/3`

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Chapter 10: Trigonometric Ratios - EXERCISE 10.3 [Page 10.37]

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R.D. Sharma Mathematics [English] Class 10
Chapter 10 Trigonometric Ratios
EXERCISE 10.3 | Q 16. (iii) | Page 10.37
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