Question

Prove that: under root secA-tanA/secA+tanA × under root cosecA-cotA/cosecA+cotA = (secA-1) (cosecA-1)

Answer 1

Correct Question:

$\sf{\sqrt{\dfrac{\sec A-\tan A}{\sec A+\tan A}}.\sqrt{\dfrac{cosec\;A -\cot A}{cosec\;A+\cot A}} = (cosec\;A-1)\;(\sec A - 1)}$

Solution:

Firstly we take LHS Part,

$\sf{\implies \sqrt{\dfrac{\sec A-\tan A}{\sec A+\tan A}}.\sqrt{\dfrac{cosec\;A -\cot A}{cosec\;A+\cot A}}}$

$\sf{\implies \sqrt{\dfrac{\sec A - \tan A}{\sec A + \tan A}\times \dfrac{\sec A - \tan A}{\sec A - \tan A}}. \sqrt{\dfrac{cosec\;A - \cot A}{cosec\;A+\cot A}\times \dfrac{cosec\;A-\cot A}{cosec\;A-\cot A}}}$

$\sf{\implies \sqrt{\dfrac{(\sec A - \tan A)^{2}}{\sec^{2} A - \tan^{2} A}}\times \sqrt{\dfrac{(cosec\;A - \cot A)^{2}}{cosec^{2} A - \cot^{2} A}}}$

We know that sec² A - tan² A = 1 & cosec² A - cot² A = 1

$\sf{\implies \sqrt{\dfrac{(\sec A - \tan A)^{2}}{1}} \times \sqrt{\dfrac{(cosec\;A - \cot A)^{2}}{1}}}$

$\sf{\implies (\sec A - \tan A)\;(cosec \;A-\cot A)}$

$\sf{\implies \bigg(\dfrac{1}{\cos A}-\dfrac{\sin A}{\cos A}\bigg)\;\bigg(\dfrac{1}{\sin A}-\dfrac{\cos A}{\sin A}\bigg)}$

$\sf{\implies \dfrac{(1-\sin A)\;(1-\cos A)}{\cos A . \sin A}}$

$\sf{\implies \bigg(\dfrac{1-\sin A}{\sin A}\bigg)\;\bigg(\dfrac{1-\cos A}{\cos A}\bigg)}$

$\sf{\implies (cosec \;A - 1)\;(\sec A - 1)}$

LHS = RHS

Hence Proved