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Mathematics Question on Trigonometric Identities

Prove the following identities, where the angles involved are acute angles for which the expressions are defined:(cosec θ−cot θ)2=(1−cos θ)(1+cos θ) (\text{cosec θ}-\text{cot θ})²=\frac{(1-\text{cos θ})}{(1 +\text{cos θ})}

Answer

(cosec θ−cot θ)2=(1−cos θ)(1+cos θ) (\text{cosec θ}-\text{cot θ})²=\frac{(1-\text{cos θ})}{(1 +\text{cos θ})}

L.H.S =(cosec θ - cot θ)2 (\text{cosec θ - cot θ})²
=(1sin θ−cos θsin θ)²= \left(\frac{1}{\text{sin θ}} - \frac{\text{cos θ}}{\text{sin θ}}\right)^²

=(1−cos θ)2(sin θ)2= \frac{(1 - \text{cos θ})²}{(\text{sin θ})²}

=(1−cos θ)2sin² θ= \frac{(1 - \text{cos θ})²}{\text{sin² θ}}

=(1−cos θ)2(1−cos²θ)=\frac{ (1 - \text{cos θ})²}{(1 - \text{cos²θ})} (By Identity sin A + cos A = 1 Hence, 1 - cos A= sin A)

=(1−cos θ)2(1−cos θ)(1+cos θ)= \frac{(1 - \text{cos θ})²}{ (1 - \text{cos θ})(1 + \text{cos θ})} [Using a² - b² = (a + b) (a - b)]

=(1− cos θ)(1+cos θ)=\frac{ (1 -\text{ cos θ})}{(1 + \text{cos θ})}
= RHS