KnowledgeBoat Logo
|

Mathematics

Prove the following identity:

(cot⁡2A(cosec⁡A+1)2)=(1−sin⁡A1+sin⁡A)\Big(\dfrac{\cot^2 A}{(\cosec A + 1)^2}\Big) = \Big(\dfrac{1 - \sin A}{1 + \sin A}\Big)

Trigonometric Identities

2 Likes

Answer

The L.H.S. of the equation can be written as,

⇒(cot⁡2A(cosec⁡A+1)2)⇒cosec⁡2A−1(cosec⁡A+1)2⇒(cosec⁡A−1)(cosec⁡A+1)(cosec⁡A+1)2⇒cosec⁡A−1(cosec⁡A+1)⇒1sin⁡A−11sin⁡A+1⇒1−sin⁡Asin⁡A1+sin⁡Asin⁡A⇒1−sin⁡A1+sin⁡A.\Rightarrow \Big(\dfrac{\cot^2 A}{(\cosec A + 1)^2}\Big) \\[1em] \Rightarrow \dfrac{\cosec^2 A - 1}{(\cosec A + 1)^2}\\[1em] \Rightarrow \dfrac{(\cosec A - 1)(\cosec A + 1)}{(\cosec A + 1)^2}\\[1em] \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}{1 + \sin A}.

Since, L.H.S. = R.H.S.

Hence, proved that (cot⁡2A(cosec⁡A+1)2)=(1−sin⁡A1+sin⁡A)\Big(\dfrac{\cot^2 A}{(\cosec A + 1)^2}\Big) = \Big(\dfrac{1 - \sin A}{1 + \sin A}\Big).

Answered By

1 Like

Related Questions