Prove the following identity:
(1−sinA1+sinA)=secA−tanA\sqrt{\Big(\dfrac{1 - \sin A}{1 + \sin A}\Big)} = \sec A - \tan A(1+sinA1−sinA)=secA−tanA
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The L.H.S. of the equation can be written as,
⇒(1−sinA1+sinA)⇒1−sinA1+sinA×1−sinA1−sinA⇒(1−sinA)21−sin2A⇒(1−sinA)2cos2A⇒(1−sinA)cosA⇒1cosA−sinAcosA⇒secA−tanA.\Rightarrow \sqrt{\Big(\dfrac{1 - \sin A}{1 + \sin A}\Big)} \\[1em] \Rightarrow \sqrt{\dfrac{1 - \sin A}{1 + \sin A} \times \dfrac{1 - \sin A}{1 - \sin A} } \\[1em] \Rightarrow \sqrt{\dfrac{(1 - \sin A)^2}{1 - \sin^2 A}} \\[1em] \Rightarrow \sqrt{\dfrac{(1 - \sin A)^2}{\cos^2 A}} \\[1em] \Rightarrow \dfrac{(1 - \sin A)}{\cos A} \\[1em] \Rightarrow \dfrac{1}{\cos A} - \dfrac{\sin A}{\cos A} \\[1em] \Rightarrow \sec A - \tan A.⇒(1+sinA1−sinA)⇒1+sinA1−sinA×1−sinA1−sinA⇒1−sin2A(1−sinA)2⇒cos2A(1−sinA)2⇒cosA(1−sinA)⇒cosA1−cosAsinA⇒secA−tanA.
Since, L.H.S. = R.H.S.
Hence, proved that (1−sinA1+sinA)=secA−tanA\sqrt{\Big(\dfrac{1 - \sin A}{1 + \sin A}\Big)} = \sec A - \tan A(1+sinA1−sinA)=secA−tanA.
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cotA−tanA=(2cos2A−1sinAcosA)\cot A - \tan A = \Big(\dfrac{2 \cos^2 A - 1}{\sin A \cos A}\Big)cotA−tanA=(sinAcosA2cos2A−1)
(1+cosA1−cosA)=cosecA+cotA\sqrt{\Big(\dfrac{1 + \cos A}{1 - \cos A}\Big)} = \cosec A + \cot A(1−cosA1+cosA)=cosecA+cotA
(cot2A(cosecA+1)2)=(1−sinA1+sinA)\Big(\dfrac{\cot^2 A}{(\cosec A + 1)^2}\Big) = \Big(\dfrac{1 - \sin A}{1 + \sin A}\Big)((cosecA+1)2cot2A)=(1+sinA1−sinA)
(cosA1−tanA)+(sin2AsinA−cosA)=cosA+sinA\Big(\dfrac{\cos A}{1 - \tan A}\Big) + \Big(\dfrac{\sin^2 A}{\sin A - \cos A}\Big) = \cos A + \sin A(1−tanAcosA)+(sinA−cosAsin2A)=cosA+sinA
