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Mathematics

Prove the following identities :

cot2A(cosec A + 1)2=1 - sin A1 + sin A\dfrac{\text{cot}^2 A}{\text{(cosec A + 1)}^2} = \dfrac{\text{1 - sin A}}{\text{1 + sin A}}

Trigonometric Identities

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Answer

By formula,

cot2 A = cosec2 A - 1

Solving L.H.S. of the equation :

⇒cot2A(cosec A + 1)2⇒cosec2A−1(cosec A + 1)2⇒(cosec A - 1)(cosec A + 1)(cosec A + 1)2⇒cosec A - 1cosec A + 1⇒1sin A−11sin A+1⇒1 - sin Asin A1 + sin Asin A⇒(1 - sin A)× sin A(1 + sin A)× sin A⇒(1 - sin A)(1 + sin A).\Rightarrow \dfrac{\text{cot}^2 A}{\text{(cosec A + 1)}^2} \\[1em] \Rightarrow \dfrac{\text{cosec}^2 A - 1}{\text{(cosec A + 1)}^2} \\[1em] \Rightarrow \dfrac{\text{(cosec A - 1)(cosec A + 1)}}{\text{(cosec A + 1)}^2} \\[1em] \Rightarrow \dfrac{\text{cosec A - 1}}{\text{cosec A + 1}} \\[1em] \Rightarrow \dfrac{\dfrac{1}{\text{sin A}} - 1}{\dfrac{1}{\text{sin A}} + 1} \\[1em] \Rightarrow \dfrac{\dfrac{\text{1 - sin A}}{\text{sin A}}}{\dfrac{\text{1 + sin A}}{\text{sin A}}} \\[1em] \Rightarrow \dfrac{\text{(1 - sin A)} \times \text{ sin A}}{\text{(1 + sin A)} \times \text{ sin A}} \\[1em] \Rightarrow \dfrac{\text{(1 - sin A)}}{\text{(1 + sin A)}}.

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

Hence, proved that cot2A(cosec A + 1)2=1 - sin A1 + sin A\dfrac{\text{cot}^2 A}{\text{(cosec A + 1)}^2} = \dfrac{\text{1 - sin A}}{\text{1 + sin A}}.

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