Prove the following identity:
cotA−tanA=(2cos2A−1sinAcosA)\cot A - \tan A = \Big(\dfrac{2 \cos^2 A - 1}{\sin A \cos A}\Big)cotA−tanA=(sinAcosA2cos2A−1)
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L.H.S. of the equation can be written as,
⇒cosAsinA−sinAcosA⇒cos2A−sin2AsinAcosA⇒cos2A−(1−cos2A)sinAcosA⇒cos2A−1+cos2AsinAcosA⇒2cos2A−1sinAcosA.\Rightarrow \dfrac{\cos A}{\sin A} - \dfrac{\sin A}{\cos A} \\[1em] \Rightarrow \dfrac{\cos^2 A - \sin^2 A}{\sin A \cos A} \\[1em] \Rightarrow \dfrac{\cos^2 A - (1 - \cos^2 A)}{\sin A \cos A} \\[1em] \Rightarrow \dfrac{\cos^2 A - 1 + \cos^2 A}{\sin A \cos A} \\[1em] \Rightarrow \dfrac{2\cos^2 A - 1}{\sin A \cos A}.⇒sinAcosA−cosAsinA⇒sinAcosAcos2A−sin2A⇒sinAcosAcos2A−(1−cos2A)⇒sinAcosAcos2A−1+cos2A⇒sinAcosA2cos2A−1.
Since, L.H.S. = R.H.S.
Hence, proved that cotA−tanA=(2cos2A−1sinAcosA)\cot A - \tan A = \Big(\dfrac{2 \cos^2 A - 1}{\sin A \cos A}\Big)cotA−tanA=(sinAcosA2cos2A−1).
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(1secA+tanA)−(1cosA)=(1cosA)−(1secA−tanA)\Big(\dfrac{1}{\sec A + \tan A}\Big) - \Big(\dfrac{1}{\cos A}\Big) = \Big(\dfrac{1}{\cos A}\Big) - \Big(\dfrac{1}{\sec A - \tan A}\Big)(secA+tanA1)−(cosA1)=(cosA1)−(secA−tanA1)
(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)
(1+cosA1−cosA)=cosecA+cotA\sqrt{\Big(\dfrac{1 + \cos A}{1 - \cos A}\Big)} = \cosec A + \cot A(1−cosA1+cosA)=cosecA+cotA
(1−sinA1+sinA)=secA−tanA\sqrt{\Big(\dfrac{1 - \sin A}{1 + \sin A}\Big)} = \sec A - \tan A(1+sinA1−sinA)=secA−tanA
