Prove the following identity:
(1−tanθ1−cotθ)2=tan2θ\Big(\dfrac{1 - \tan \theta}{1 - \cot \theta}\Big)^2 = \tan^2 \theta(1−cotθ1−tanθ)2=tan2θ
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The L.H.S of above equation can be written as,
⇒(1−tanθ1−cotθ)2⇒(1−tanθ1−1tanθ)2⇒((1−tanθ)(tanθ)tanθ−1)2⇒((1−tanθ)(tanθ)−(1−tanθ))2⇒(−tanθ)2⇒tan2θ.\Rightarrow \Big(\dfrac{1 - \tan \theta}{1 - \cot \theta}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{1 - \tan \theta}{1 - \dfrac{1}{\tan \theta}}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{(1 - \tan \theta)(\tan \theta)}{{\tan \theta - 1}}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{(1 - \tan \theta)(\tan \theta)}{{-(1 - \tan \theta)}}\Big)^2 \\[1em] \Rightarrow (-\tan \theta)^2 \\[1em] \Rightarrow \tan^2 \theta .⇒(1−cotθ1−tanθ)2⇒(1−tanθ11−tanθ)2⇒(tanθ−1(1−tanθ)(tanθ))2⇒(−(1−tanθ)(1−tanθ)(tanθ))2⇒(−tanθ)2⇒tan2θ.
Since, L.H.S. = R.H.S.
Hence, proved that (1−tanθ1−cotθ)2=tan2θ\Big(\dfrac{1 - \tan \theta}{1 - \cot \theta}\Big)^2 = \tan^2 \theta(1−cotθ1−tanθ)2=tan2θ.
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(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
(cosA1+sinA)+tanA=secA\Big(\dfrac{\cos A}{1 + \sin A}\Big) + \tan A = \sec A(1+sinAcosA)+tanA=secA
(sinθ+cosθ)(tanθ+cotθ)=secθ+cosecθ(\sin \theta + \cos \theta)(\tan \theta + \cot \theta) = \sec \theta + \cosec \theta(sinθ+cosθ)(tanθ+cotθ)=secθ+cosecθ
