Question

Prove that

\[\frac{cosec(90^\circ + x) + \cot(450^\circ + x)}{cosec(90^\circ - x) + \tan(180^\circ - x)} + \frac{\tan(180^\circ + x) + \sec(180^\circ - x)}{\tan(360^\circ + x) - \sec( - x)} = 2\]

 

Answer 1

LHS = \[\frac{cosec \left( 90^\circ + x \right) + \cot \left( 450^\circ + x \right)}{cosec \left( 90^\circ - x \right) + \tan \left( 180^\circ - x \right)} + \frac{\tan \left( 180^\circ + x \right) + \sec \left( 180^\circ - x \right)}{\tan \left( 360^\circ + x \right) - \sec \left( - x \right)}\]
\[ = \frac{cosec\left( 90^\circ + x \right) + \cot\left( 450^\circ + x \right)}{cosec \left( 90^\circ - x \right) + \tan\left( 180^\circ - x \right)} + \frac{\tan \left( 180^\circ + x \right) + \sec \left( 180^\circ - x \right)}{\tan \left( 360^\circ + x \right) - \sec \left( - x \right)}\]
\[ = \frac{cosec\left( 90^\circ + x \right) + \cot \left( 90^\circ \times 5 + x \right)}{cosec\left( 90^\circ - x \right) + \tan \left( 90^\circ \times 2 - x \right)} + \frac{\tan \left( 90^\circ \times 2 + x \right) + \sec \left( 90^\circ \times 2 - x \right)}{\tan\left( 90^\circ \times 4 + x \right) - \sec\left( - x \right)}\]
\[ = \frac{\sec x + \cot \left( 90^\circ \times 5 + x \right)}{cosec\left( 90^\circ- x \right) + \tan \left( 90^\circ \times 2 - x \right)} + \frac{\tan \left( 90^\circ \times 2 + x \right) + \sec \left( 90^\circ \times 2 - x \right)}{\tan \left( 90^\circ \times 4 + x \right) - \sec \left( - x \right)}\]
\[ = \frac{\sec x - \tan x}{\sec x - \tan x} + \frac{\tan x - \sec x}{\tan x - \sec x}\]
\[ = 1 + 1\]
\[ = 2\]
 = RHS
Hence proved.