Question

A.


`\cot^(2) - \cos ^2a = \cot ^(2) a * \cos ^(2) a`
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Answer 1

Explanation:

We need to prove that,

`\cot^2 a-\cos^2 a=\cot^2 a* \cos^2 a`

Taking LHS,

`\cot^2 a-\cos^2 a`

We know that,

`\cot a=(\cos a)/(\sin a)`

​
`=((\cos a)/(\sin a))^2 -\cos^2 a\\\\=\cos^2 a((1)/(\sin^2 a)-1)\\\\=\cos^2 a((1-\sin^2a)/(\sin^2 a))\\\\=\cos^2 a* (\cos^2 a)/(\sin^2 a)\\\\=(\cos ^2 a)/(\sin^2 a)* \cos^2a\\\\\because \ (\cos ^2 a)/(\sin^2 a)=\cot^2 a\\\\=\cot^2 a * \cos^2a\\\\=RHS`

So, LHS = RHS

Hence, proved.