Question

prove that √sec-1/sec+1 +√sec+1/sec-1 =2 cosec

Answer 1

hope it's clear............

Answer 2

Answer:

$\sqrt{\frac{sec\theta-1}{sec\theta+1}}+\sqrt{\frac{sec\theta+1}{sec\theta-1}}=2cosec\theta$

Step-by-step explanation:

$LHS = \sqrt{\frac{sec\theta-1}{sec\theta+1}}+\sqrt{\frac{sec\theta+1}{sec\theta-1}}\\=\frac{\sqrt{(sec\theta-1)^{2}}+\sqrt{(sec\theta+1)^{2}}}{\sqrt{(sec\theta+1)(sec\theta-1)}}$

$=\frac{sec\theta-1+sec\theta+1}{\sqrt{sec^{2}\theta-1^{2}}}$

$=\frac{2sec\theta}{\sqrt{tan^{2}\theta}}$

/* By Trigonometric identity:

sec²A-1 = tan²A*/

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

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

Therefore,

$\sqrt{\frac{sec\theta-1}{sec\theta+1}}+\sqrt{\frac{sec\theta+1}{sec\theta-1}}=2cosec\theta$

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