Prove the following identity:
(tanθ+sinθtanθ−sinθ)=(secθ+1secθ−1)\Big(\dfrac{\tan \theta + \sin \theta}{\tan \theta - \sin \theta}\Big) = \Big(\dfrac{\sec \theta + 1}{\sec \theta - 1}\Big)(tanθ−sinθtanθ+sinθ)=(secθ−1secθ+1)
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Solving L.H.S of the equation,
⇒tanθ+sinθtanθ−sinθ⇒sinθcosθ+sinθsinθcosθ−sinθ⇒sinθ(1cosθ+1)sinθ(1cosθ−1)⇒1cosθ+11cosθ−1⇒secθ+1secθ−1.\Rightarrow \dfrac{\tan \theta + \sin \theta}{\tan \theta - \sin \theta} \\[1em] \Rightarrow \dfrac{\dfrac{\sin \theta}{\cos \theta} + \sin \theta}{\dfrac{\sin \theta}{\cos \theta} - \sin \theta} \\[1em] \Rightarrow \dfrac{\sin \theta \Big(\dfrac{1}{\cos \theta} + 1\Big)}{\sin \theta \Big(\dfrac{1}{\cos \theta} - 1\Big)} \\[1em] \Rightarrow \dfrac{\dfrac{1}{\cos \theta} + 1}{\dfrac{1}{\cos \theta} - 1} \\[1em] \Rightarrow \dfrac{\sec \theta + 1}{\sec \theta - 1}.⇒tanθ−sinθtanθ+sinθ⇒cosθsinθ−sinθcosθsinθ+sinθ⇒sinθ(cosθ1−1)sinθ(cosθ1+1)⇒cosθ1−1cosθ1+1⇒secθ−1secθ+1.
Since, L.H.S. = R.H.S.
Hence, proved that (tanθ+sinθtanθ−sinθ)=(secθ+1secθ−1)\Big(\dfrac{\tan \theta + \sin \theta}{\tan \theta - \sin \theta}\Big) = \Big(\dfrac{\sec \theta + 1}{\sec \theta - 1}\Big)(tanθ−sinθtanθ+sinθ)=(secθ−1secθ+1).
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(tanA1−cotA)+(cotA1−tanA)=secAcosecA+1\Big(\dfrac{\tan A}{1 - \cot A}\Big) + \Big(\dfrac{\cot A}{1 - \tan A}\Big) = \sec A \cosec A + 1(1−cotAtanA)+(1−tanAcotA)=secAcosecA+1
(sinA1+cotA)−(cosA1+tanA)=sinA−cosA\Big(\dfrac{\sin A}{1 + \cot A}\Big) - \Big(\dfrac{\cos A}{1 + \tan A}\Big) = \sin A - \cos A(1+cotAsinA)−(1+tanAcosA)=sinA−cosA
(cotθ+cosecθ−1cotθ−cosecθ+1)=(1+cosθsinθ)\Big(\dfrac{\cot \theta + \cosec \theta - 1}{\cot \theta - \cosec \theta + 1}\Big) = \Big(\dfrac{1 + \cos \theta}{\sin \theta}\Big)(cotθ−cosecθ+1cotθ+cosecθ−1)=(sinθ1+cosθ)
(cotA−12−sec2A)=(cotA1+tanA)\Big(\dfrac{\cot A - 1}{2 - \sec^2 A}\Big) = \Big(\dfrac{\cot A}{1 + \tan A}\Big)(2−sec2AcotA−1)=(1+tanAcotA)
