Prove the following identity:
(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)
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Solving L.H.S:
⇒secA−1secA+1⇒1cosA−11cosA+1⇒1−cosAcosA1+cosAcosA⇒(1−cosA)×cosA(1+cosA)×cosA⇒1−cosA1+cosA.\Rightarrow \dfrac{\sec A - 1}{\sec A + 1} \\[1em] \Rightarrow \dfrac{\dfrac{1}{\cos A} - 1}{\dfrac{1}{\cos A} + 1} \\[1em] \Rightarrow \dfrac{\dfrac{1 - \cos A}{\cos A}}{\dfrac{1 + \cos A}{\cos A}} \\[1em] \Rightarrow \dfrac{(1 - \cos A) \times \cos A}{(1 + \cos A) \times \cos A} \\[1em] \Rightarrow \dfrac{1 - \cos A}{1 + \cos A}.⇒secA+1secA−1⇒cosA1+1cosA1−1⇒cosA1+cosAcosA1−cosA⇒(1+cosA)×cosA(1−cosA)×cosA⇒1+cosA1−cosA.
Since, L.H.S. = R.H.S.
Hence, proved that (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).
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(cosecA+1cosecA−1)=(1+sinA1−sinA)\Big(\dfrac{\cosec A + 1}{\cosec A - 1}\Big) = \Big(\dfrac{1 + \sin A}{1 - \sin A}\Big)(cosecA−1cosecA+1)=(1−sinA1+sinA)
(sinA×tanA1−cosA)=1+secA\Big(\dfrac{\sin A \times \tan A}{1 - \cos A}\Big) = 1 + \sec A(1−cosAsinA×tanA)=1+secA
(1tanA+cotA)=cosA×sinA\Big(\dfrac{1}{\tan A + \cot A}\Big) = \cos A \times \sin A(tanA+cotA1)=cosA×sinA
(1 + tan A)2 + (1 - tan A)2 = 2 sec2 A
