3 years ago

How do I establish this identity??

1 answer:

8 0

$\frac{cosec\theta+cotan\theta} {sec\theta+tan\theta}=\frac{cosec\theta+cotan\theta} {sec\theta+tan\theta}\times\frac{cosec\theta-cotan\theta}{cosec\theta-cotan\theta}\times\frac{sec\theta-tan\theta} {sec\theta-tan\theta}$

And use that,

$\sec^2\theta-\tan^2\theta=1$

$cosec^2\theta-cotan^2\theta=1$

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