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

\therefore LHS = \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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