Question

PROVE THE FOLLOWING IDENTITY

Answer 1

Answer:

See below.

Step-by-step explanation:

LHS:

$\frac{1}{cosec\Theta + cot\Theta } - \frac{1}{sin\Theta}$

$=\frac{1}{\frac{1}{sin\Theta}+\frac{cos\Theta}{sin\Theta}} - \frac{1}{sin\Theta}$              $[cosec\Theta = \frac{1}{sin\Theta}$ $and$ $cot\Theta = \frac{cos\Theta}{sin\Theta}]$

$=\frac{sin\Theta}{1+cos\Theta} - \frac{1}{sin\Theta}$

$=\frac{(sin\Theta \times sin\Theta) - 1 \times (1+cos\Theta)}{(1+cos\Theta) \times sin\Theta}$

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

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

$=\frac{-cos^2\Theta - cos\Theta}{sin\Theta (1+cos\Theta)}$                       $[sin^2\Theta + cos^2\Theta = 1]$

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

$=-cot\Theta$

RHS:

$\frac{1}{sin\Theta} - \frac{1}{cosec\Theta - cot\Theta}$

$=\frac{1}{sin\Theta} - \frac{1}{\frac{1}{sin\Theta}-\frac{cos\Theta}{sin\Theta}}$               $[cosec\Theta = \frac{1}{sin\Theta}$ $and$ $cot\Theta = \frac{cos\Theta}{sin\Theta}]$

$=\frac{1}{sin\Theta}- \frac{sin\Theta}{1-cos\Theta}$

$=\frac{ 1 \times (1-cos\Theta) - (sin\Theta \times sin\Theta)}{sin\Theta \times (1-cos\Theta)}$

$=\frac{(1-cos\Theta)-(sin^2\Theta) }{sin\Theta (1-cos\Theta)}$

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

$=\frac{cos^2\Theta - cos\Theta}{sin\Theta (1-cos\Theta)}$                        $[sin^2\Theta + cos^2\Theta = 1]$

$=\frac{-cos\Theta(1-cos\Theta)}{sin\Theta (1-cos\Theta)}$

$=-cot\Theta$

Therefore LHS = RHS. Hence Proved.

Any corrections or suggestions for improvement are welcome :)