Question

2 sec square theta - sec^4 theta -2 cosec square theta + cosec^4 theta = cot^4 - tan^4 theta?)Prove.

Answer 1

Answer:

$2sec^{2}\theta-sec^{4}\theta-2cosec^{2}\theta+cosec^{4}\theta\\=cot^{4}\theta-tan^{4}\theta$

Step-by-step explanation:

$LHS=2sec^{2}\theta-sec^{4}\theta-2cosec^{2}\theta+cosec^{4}\theta\\=2sec^{2}\theta-2cosec^{2}\theta-sec^{4}\theta+cosec^{4}\theta\\=2(sec^{2}\theta-cosec^{2}\theta)+(cosec^{4}\theta-sec^{4}\theta)\\=2(sec^{2}\theta-cosec^{2}\theta)+(cosec^{2}\theta)^{2}-(sec^{2}\theta)^{2})\\=2(sec^{2}\theta-cosec^{2}\theta)+(cosec^{2}\theta-sec^{2}\theta)(cosec^{2}\theta+sec^{2}\theta)\\=(sec^{2}\theta-cosec^{2}\theta)[2-cosec^{2}\theta-sec^{2}\theta]\\=[1+tan^{2}\theta-(1+cot^{2}\theta][(1-cosec^{2}\theta)+(1-sec^{2}\theta)]\\=(1+tan^{2}\theta-1-cot^{2}\theta)(-cot^{2}\theta-tan^{2}\theta)\\=(tan^{2}\theta-cot^{2}\theta)[-(tan^{2}\theta+cot^{2}\theta)]\\=-(tan^{4}\theta-cot^{4}\theta)\\=cot^{4}\theta-tan^{4}\theta\\=RHS$

Therefore,

$2sec^{2}\theta-sec^{4}\theta-2cosec^{2}\theta+cosec^{4}\theta\\=cot^{4}\theta-tan^{4}\theta$

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Answer 2

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