Prove that root[(sec A - 1)/(sec A + 1)] + root[(sec A + 1)/(sec A

Q) Prove that:

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

Ans: Let’s start from LHS

LHS = $\sqrt \frac {\sec A - 1}{\sec A + 1} + \sqrt \frac {\sec A + 1}{\sec A - 1}$

Since sec A = $\frac{1}{\cos A}$

$\therefore$ LHS = $\sqrt \frac {\frac{1}{\cos A} - 1}{\frac{1}{\cos A} + 1} + \sqrt \frac {\frac{1}{\cos A} + 1}{\frac{1}{\cos A} - 1}$

= $\sqrt \frac {\frac{1 - \cos A}{\cos A}}{\frac{1 + \cos A}{\cos A}} + \sqrt \frac {\frac{1 + \cos A}{\cos A}}{\frac{1 - \cos A}{\cos A}}$

= $\sqrt \frac {(1 - \cos A)}{(1 + \cos A)} + \sqrt \frac {(1 + \cos A)}{(1 - \cos A)}$

= $\sqrt \frac {(1 - \cos A)(1 + \cos A)}{(1 + \cos A)(1 + \cos A)} + \sqrt \frac {(1 + \cos A)(1 - \cos A)}{(1 - \cos A)(1 - \cos A)}$

= $\sqrt \frac {(1 - \cos^2 A)}{(1 + \cos A)^2} + \sqrt \frac {(1 - \cos^2 A)}{(1 - \cos A)^2}$

= $\sqrt \frac {\sin^2 A}{(1 + \cos A)^2} + \sqrt \frac {\sin^2 A}{(1 - \cos A)^2}$

= $\frac {\sin A}{(1 + \cos A)} + \frac {\sin A}{(1 - \cos A)}$

= $\frac {(\sin A)(1 - \cos A) + (\sin A)(1 + \cos A)}{(1 + \cos A)(1 - \cos A)}$

= $\frac {2 \sin A}{(1 - \cos^2 A)}$

= $\frac {2 \sin A}{(\sin^2 A)}$

= $\frac {2}{(\sin A)}$

= 2 cosec $A$

= RHS …………… Hence Proved

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