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महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता ११ वी

Prove the following: 2 sec2θ – sec4θ – 2cosec2θ + cosec4θ = cot4θ – tan4θ

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प्रश्न

Prove the following:

2 sec2θ – sec4θ – 2cosec2θ + cosec4θ = cot4θ – tan4θ

बेरीज
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उत्तर

L.H.S. = 2 sec2θ – sec4θ – 2cosec2θ + cosec4θ

= 2 sec2θ – (sec2θ)2 – 2cosec2θ + (cosec2θ)2

= 2(1 + tan2θ) – (1 + tan2θ)2 – 2(1 + cot2θ) + (1 + cot2θ)2 

= 2 + 2tan2θ – (1 + 2tan2θ + tan4θ) – 2 – 2cot2θ + 1 + 2cot2θ + cot4θ

= 2 + 2tan2θ – 1 – 2tan2θ – tan4θ – 2 – 2cot2θ + 1 + 2cot2θ + cot4θ

= cot4θ – tan4θ

= R.H.S.

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पाठ 2: Trigonometry - 1 - MISCELLANEOUS EXERCISE - 2 [पृष्ठ ३३]

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बालभारती Mathematics and Statistics (Arts and Science) Part 1 [English] Standard 11 Maharashtra State Board
पाठ 2 Trigonometry - 1
MISCELLANEOUS EXERCISE - 2 | Q 10) iv) | पृष्ठ ३३

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