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Mathematics

Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:

1 - cos A1 + cos A=cosec A - cot A\sqrt{\dfrac{\text{1 - cos A}}{\text{1 + cos A}}} = \text{cosec A - cot A}

Trigonometric Identities

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Answer

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

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

Since, L.H.S. = R.H.S. hence, proved that 1 - cos A1 + cos A\sqrt{\dfrac{\text{1 - cos A}}{\text{1 + cos A}}} = cosec A - cot A.

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