# Prove that C O S E C ( 90 ∘ + X ) + Cot ( 450 ∘ + X ) C O S E C ( 90 ∘ − X ) + Tan ( 180 ∘ − X ) + Tan ( 180 ∘ + X ) + Sec ( 180 ∘ − X ) Tan ( 360 ∘ + X ) − Sec ( − X ) = 2 - Mathematics

Prove that

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

#### Solution

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.

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 5 Trigonometric Functions
Exercise 5.3 | Q 3.2 | Page 39