Question

2 sec square minus 4 minus cos squared + cos 4A​

Answer 1

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Step-by-step explanation:

Answer 2

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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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