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