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Mathematics

Prove that :

(1 + cos Asin A)2=1 + cos A1 - cos A\Big(\dfrac{\text{1 + cos A}}{\text{sin A}}\Big)^2 = \dfrac{\text{1 + cos A}}{\text{1 - cos A}}

Trigonometric Identities

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Answer

To prove:

Equation : (1 + cos Asin A)2=1 + cos A1 - cos A\Big(\dfrac{\text{1 + cos A}}{\text{sin A}}\Big)^2 = \dfrac{\text{1 + cos A}}{\text{1 - cos A}}.

Solving L.H.S. of the above equation :

⇒(1 + cos Asin A)2⇒(1 + cos A)2sin2A⇒(1 + cos A)2(1 - cos2A)⇒(1 + cos A)2(1 - cos A)(1 + cos A)⇒1 + cos A1 - cos A.\Rightarrow \Big(\dfrac{\text{1 + cos A}}{\text{sin A}}\Big)^2 \\[1em] \Rightarrow \dfrac{(\text{1 + cos A})^2}{\text{sin}^2 A} \\[1em] \Rightarrow \dfrac{(\text{1 + cos A})^2}{(\text{1 - cos}^2 A)} \\[1em] \Rightarrow \dfrac{(\text{1 + cos A})^2}{\text{(1 - cos A)(1 + cos A)}} \\[1em] \Rightarrow \dfrac{\text{1 + cos A}}{\text{1 - cos A}}.

∴ L.H.S. = R.H.S.

Hence, proved that (1 + cos Asin A)2=1 + cos A1 - cos A\Big(\dfrac{\text{1 + cos A}}{\text{sin A}}\Big)^2 = \dfrac{\text{1 + cos A}}{\text{1 - cos A}}.

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