Prove the following identity:
(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)
9 Likes
Solving L.H.S:
⇒cosecA+1cosecA−1⇒1sinA+11sinA−1⇒1+sinAsinA1−sinAsinA⇒(1+sinA)×sinA(1−sinA)×sinA⇒1+sinA1−sinA.\Rightarrow \dfrac{\cosec A + 1}{\cosec A - 1} \\[1em] \Rightarrow \dfrac{\dfrac{1}{\sin A} + 1}{\dfrac{1}{\sin A} - 1} \\[1em] \Rightarrow \dfrac{\dfrac{1 + \sin A}{\sin A}}{\dfrac{1 - \sin A}{\sin A}} \\[1em] \Rightarrow \dfrac{(1 + \sin A) \times \sin A}{(1 - \sin A) \times \sin A} \\[1em] \Rightarrow \dfrac{1 + \sin A}{1 - \sin A}.⇒cosecA−1cosecA+1⇒sinA1−1sinA1+1⇒sinA1−sinAsinA1+sinA⇒(1−sinA)×sinA(1+sinA)×sinA⇒1−sinA1+sinA.
Since, L.H.S. = R.H.S.
Hence, proved that (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).
Answered By
4 Likes
(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
(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
