Prove the following identity:
(1tanA+cotA)=cosA×sinA\Big(\dfrac{1}{\tan A + \cot A}\Big) = \cos A \times \sin A(tanA+cotA1)=cosA×sinA
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Solving L.H.S. of the equation :
⇒1tanA+cotA⇒1sinAcosA+cosAsinA⇒1sin2A+cos2AsinAcosA⇒sinAcosAsin2A+cos2A⇒sinAcosA\Rightarrow \dfrac{1}{\tan A + \cot A} \\[1em] \Rightarrow \dfrac{1}{\dfrac{\sin A}{\cos A} + \dfrac{\cos A}{\sin A}} \\[1em] \Rightarrow \dfrac{1}{\dfrac{\sin^2 A + \cos^2 A}{\sin A \cos A}} \\[1em] \Rightarrow \dfrac{\sin A \cos A}{\sin^2 A + \cos^2 A} \\[1em] \Rightarrow \sin A \cos A⇒tanA+cotA1⇒cosAsinA+sinAcosA1⇒sinAcosAsin2A+cos2A1⇒sin2A+cos2AsinAcosA⇒sinAcosA
Since, L.H.S. = R.H.S.
Hence, proved that (1tanA+cotA)=cosA×sinA\Big(\dfrac{1}{\tan A + \cot A}\Big) = \cos A \times \sin A(tanA+cotA1)=cosA×sinA.
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(secA−1secA+1)=(1−cosA1+cosA)\Big(\dfrac{\sec A - 1}{\sec A + 1}\Big) = \Big(\dfrac{1 - \cos A}{1 + \cos A}\Big)(secA+1secA−1)=(1+cosA1−cosA)
(sinA×tanA1−cosA)=1+secA\Big(\dfrac{\sin A \times \tan A}{1 - \cos A}\Big) = 1 + \sec A(1−cosAsinA×tanA)=1+secA
(1 + tan A)2 + (1 - tan A)2 = 2 sec2 A
(sin2 θ - 1) (tan2 θ + 1) + 1 = 0
