# Prove the Following Identities ( 1 + Cot X + Tan X ) ( Sin X − Cos X ) Sec 3 X − C O S E C 3 X = Sin 2 X Cos 2 X - Mathematics

Prove the following identities

$\frac{\left( 1 + \cot x + \tan x \right) \left( \sin x - \cos x \right)}{\sec^3 x - {cosec}^3 x} = \sin^2 x \cos^2 x$

#### Solution

$\frac{\left( 1 + \cot x + \tan x \right) \left( \sin x - \cos x \right)}{\sec^3 x - {cosec}^3 x} = \sin^2 x \cos^2 x$
$\text{LHS} = \frac{\left( 1 + \cot x + \tan x \right)\left( \sin x - \cos x \right)}{\sec^3 x - {cosec}^3 x}$
$= \frac{\left( 1 + \frac{\cos x}{\sin x} + \frac{\sin x}{\cos x} \right)\left( \sin x - \cos x \right)}{\frac{1}{\cos^3 x} - \frac{1}{\sin^3 x}}$
$= \frac{\left( \sin x \cos x + \cos^2 x + \sin^2 x \right)\left( \sin x - \cos x \right)\left( \sin^2 x \cos^2 x \right)}{\left( \sin^3 x - \cos^3 x \right)}$
$= \frac{\left( 1 + \sin x \cos x \right)\left( \sin x - \cos x \right)\left( \sin^2 x \cos^2 x \right)}{\left( \sin x - \cos x \right)\left( \sin^2 x + \cos^2 x + \sin x \cos x \right)}$
$= \sin^2 x \cos^2 x$
= 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 5.1 | Q 14 | Page 18