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महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता ११ वी

Prove the following: sec 840° . cot (– 945°) + sin 600° tan (– 690°) = 32 - Mathematics and Statistics

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प्रश्न

Prove the following:

sec 840° . cot (– 945°) + sin 600° tan (– 690°) = `3/2`

बेरीज
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उत्तर

sec 840° = sec (720° + 120°)

= sec (2 x 360° + 120°)

= sec (120°)

= sec (90° + 30°)

= – cosec 30°

= – 2

cot (– 945°) = – cot 945°

= – cot (720° + 225°)

= – cot (2 x 360° + 225°)

= – cot (225°)

= – cot (180° + 45°)

= – cot 45°

= – 1

sin 600° = sin (360° + 240°)

= sin (240°)

= sin (180° + 60°)

= – sin 60°

= `-sqrt(3)/2`

tan (– 690°) = – tan 690°

= – tan (360° + 330°)

= – tan (330°)

= – tan (360° – 30°)

= – (– tan – 30°)

= tan 30°

= `1/sqrt(3)`

L.H.S. = sec 840° cot (– 945°) + sin 600° tan (– 690°)

= `(-2) (-1) + (-sqrt(3)/2)(1/sqrt(3))`

= `2 - 1/2`

= `3/2`

= R.H.S.

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पाठ 3: Trigonometry - 2 - Exercise 3.2 [पृष्ठ ४२]

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