Question

please someone solve this. ​

Answer 1

Question:-

Prove the identity:-

• $\sf{\dfrac{Sin\theta}{Cot\theta + Cosec\theta} = 2 + \dfrac{Sin\theta}{Cot\theta - Cosec\theta}}$

Solution:-

Taking LHS,

$\sf{\dfrac{Sin\theta}{Cot\theta + Cosec\theta}}$

We already know:-

• $\sf{Cot\theta = \dfrac{Cos\theta}{Sin\theta}}$

• $\sf{Cosec\theta = \dfrac{1}{Sin\theta}}$

Hence,

$= \sf{\dfrac{Sin\theta}{\dfrac{Cos\theta}{Sin\theta} + \dfrac{1}{Sin\theta}}}$

$= \sf{\dfrac{Sin\theta}{\dfrac{Cos\theta + 1}{Sin\theta}}}$

$= \sf{\dfrac{Sin\theta \times Sin\theta}{Cos\theta + 1}}$

$= \sf{\dfrac{Sin^2\theta}{Cos\theta + 1}}$

We also know:-

• Sin²θ + Cos²θ = 1
• => Sin²θ = 1 - Cos²θ

Hence,

$= \sf{\dfrac{1 - Cos^2\theta}{1 + Cos\theta}}$

Applying the identity a² - b² = (a + b)(a - b):-

$= \sf{\dfrac{(1 + Cos\theta)(1 - Cos\theta)}{1 + Cos\theta}}$

$= \sf{1 - Cos\theta}$

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Taking RHS:-

$\sf{2 + \dfrac{Sin\theta}{Cot\theta - Cosec\theta}}$

$=\sf{2 + \dfrac{Sin\theta}{\dfrac{Cos\theta}{Sin\theta} - \dfrac{1}{Sin\theta}}}$

$=\sf{2 + \dfrac{Sin\theta}{\dfrac{Cos\theta - 1}{Sin\theta}}}$

$=\sf{2 + \dfrac{Sin\theta \times Sin\theta}{Cos\theta - 1}}$

$= \sf{2 - \bigg(\dfrac{Sin^2\theta}{1 - Cos\theta}\bigg)}$

$= \sf{2 - \bigg(\dfrac{1 - Cos^2\theta}{1 - Cos\theta}\bigg)}$

$= \sf{2 - \bigg(\dfrac{(1 - Cos\theta)(1 + Cos\theta)}{1 - Cos\theta}\bigg)}$

$= \sf{2 - (1 + Cos\theta)}$

$= \sf{2 - 1 - Cos\theta}$

$= \sf{1 - Cos\theta}$

Hence, LHS = RHS (Proved)

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