Question

1 + cos theta by sin theta minus sin theta by 1 + cos theta is equals to 2 cot theta

Answer 1

Answer:

here is your answer mate... in the form of a picture

Answer 2

Answer with Step-by-step explanation:

LHS

$\frac{1+cos\theta}{sin\theta}-\frac{sin\theta}{1+cos\theta}$

$\frac{(1+cos\theta)^2-sin^2\theta}{sin\theta(1+cos\theta)}$

$\frac{(1+cos\theta)^2-(1-cos^2\theta)}{sin\theta(1+cos\theta)}$

$sin^2\theta=1-cos^2\theta$

$\frac{(1+cos\theta)(1+cos\theta)-(1+cos\theta)(1-cos\theta)}{sin\theta(1+cos\theta)}$

Using the formula

$a^2-b^2=(a+b)(a-b)$

$\frac{(1+cos\theta)(1+cos\theta-1+cos\theta)}{sin\theta(1+cos\theta)}$

$\frac{2cos\theta}{sin\theta}$

$cot\theta=\frac{cos\theta}{sin\theta}$

$2cot\theta$

LHS=RHS

Hence proved.