Question

prove it
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Answer 1

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Answer 2

Answer:

To prove:  $\frac{\sqrt{Sec\theta-1}}{\sqrt{Sec\theta+1}}+\frac{\sqrt{Sec\theta+1}}{\sqrt{Sec\theta-1}}=2cosec\theta$

Proof,

LHS

$=\frac{\sqrt{Sec\theta-1}}{\sqrt{Sec\theta+1}}+\frac{\sqrt{Sec\theta+1}}{\sqrt{Sec\theta-1}}$

$=\frac{\sqrt{Sec\theta-1}\times\sqrt{Sec\theta-1}+\sqrt{Sec\theta+1}\times\sqrt{Sec\theta+1}}{\sqrt{Sec\theta+1}\times\sqrt{Sec\theta-1}}$$=\frac{(\sqrt{Sec\theta-1})^2+(\sqrt{Sec\theta+1})^2}{\sqrt{(Sec\theta+1)(Sec\theta-1)}}$

$=\frac{(Sec\theta-1)+(Sec\theta+1)}{\sqrt{Sec^2\theta-1^2}}$   (using identity ( x - y ) ( x + y ) = x² - y² )

$=\frac{Sec\theta-1+Sec\theta+1}{\sqrt{tan^2\theta}}$   (using identity $sec^2\theta-1=tan^2\theta$)

$=\frac{2Sec\theta}{tan\theta}$

$=\frac{2\frac{1}{cos\theta}}{\frac{sin\theta}{cos\theta}}$

$=\frac{2}{sin\theta}$

$=2\,cosec\theta$

= RHS

Hence Proved