Question

Prove the given sum​

Answer 1

$\huge{\red{\underline{\underline{\rm{Answer:}}}}}$

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

$\sf{Now,\;we\;choose\;LHS,}$

$\sf{\implies \sqrt{\dfrac{cosec\;x-1}{cosec\;x+1}}}$

$\sf{Now,\;we\;will\;do\;rationalize,}$

$\sf{\implies \sqrt{\dfrac{cosec\;x-1}{cosec\;x+1}\times \dfrac{cosec\;x-1}{cosec\;x-1}}}$

$\sf{\implies \sqrt{\dfrac{(cosec\;x-1)^{2}}{cosec^{2}\;x-1}}}$

$\sf{\implies \sqrt{\dfrac{(cosec\;x-1)^{2}}{\cot^{2}\;x}}}$

$\sf{\implies \dfrac{cosec\;x-1}{\cot x}}$

$\sf{\implies \dfrac{cosec\;x}{\cot x}-\dfrac{1}{\cot x}}$

$\sf{Now,\;as\;we\;know,}$

$\sf{\implies cosec\;x=\Bigg(\dfrac{1}{\sin x}\Bigg)}$

$\sf{\implies \cot x = \dfrac{1}{\tan x}=\dfrac{1}{\frac{\sin x}{\cos x}} =\dfrac{\cos x}{\sin x}}$

$\sf{\implies \dfrac{1}{\cot x}=\tan x}$

$\sf{So,\;now\; \dfrac{\Bigg(\dfrac{1}{\sin x}\Bigg)}{\Bigg(\dfrac{\cos x}{\sin x}\Bigg)}-\tan x}$

$\sf{\implies \dfrac{1}{\cos x}-\tan x}$

$\sf{\implies \sec x -\tan x}$

$\sf{Now,\;again\;rationalize\;it.}$

$\sf{\implies \sec x - \tan x \times \dfrac{\sec x + \tan x}{\sec x + \tan x}}$

$\sf{\implies \dfrac{\sec^{2}x-\tan^{2}x}{\sec x + \tan x}}$

$\sf{As\;we\;know,\;\sec^{2}x-\tan^{2}x=1}$

$\sf{\implies \dfrac{1}{\sec x + \tan x}}$

$\large{\red{\sf{LHS = RHS\;\;\;\;\;[Hence\;Proved]}}}$

$\sf{2).\;\dfrac{1}{\sin A+\cos A+1}+\dfrac{1}{\sin A+\cos A-1}=\sec A+cosec\;A}$

$\sf{Now,\;we\;will\;take\;LHS,}$

$\sf{\implies \dfrac{1}{\sin A+\cos A+1}+\dfrac{1}{\sin A+\cos A-1}}$

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

$\sf{\implies \dfrac{2\sin A+2\cos A}{\sin^{2}A+\sin A.\cos A-\sin A+\sin A.\cos A+\cos^{2}A-\cos A+\sin A+\cos A-1}}$

$\sf{\implies \dfrac{2(\sin A+\cos A)}{\sin^{2} A +\sin A.\cos A+\sin A.\cos A+\cos^{2}A-1}}$

$\sf{As\;we\;know,\;sin^{2}A+\cos^{2} A = 1.\;So,}$

$\sf{\implies \dfrac{2(\sin A+\cos A)}{1+2\sin A.\cos A-1}}$

$\sf{\implies \dfrac{2(\sin A+\cos A)}{2\sin A.\cos A}}$

$\sf{\implies \dfrac{\sin A+\cos A}{\sin A.\cos A}}$

$\sf{\implies \dfrac{\sin A}{\sin A.\cos A} +\dfrac{\cos A}{\sin A.\cos A}}$

$\sf{As\;we\;know\;\dfrac{1}{\cos A}=\sec A\;and,\;\dfrac{1}{\sin A}=cosec\;A}$

$\sf{\implies \sec A + cosec\;A}$

$\large{\red{\sf{LHS = RHS\;\;\;\;\;[Hence\;Proved]}}}$