# Prove the Following Identities Tan 3 X 1 + Tan 2 X + Cot 3 X 1 + Cot 2 X = 1 − 2 Sin 2 X Cos 2 X Sin X Cos X - Mathematics

Prove the following identities

$\frac{\tan^3 x}{1 + \tan^2 x} + \frac{\cot^3 x}{1 + \cot^2 x} = \frac{1 - 2 \sin^2 x \cos^2 x}{\sin x \cos x}$

#### Solution

$\text{ LHS }= \frac{\tan^3 x}{1 + \tan^2 x} + \frac{\cot^3 x}{1 + \cot^2 x}$
$= \frac{\tan^3 x}{\sec^2 x} + \frac{\cot^3 x}{{cosec}^2 x}$
$= \frac{\frac{\sin^3 x}{\cos^3 x}}{\frac{1}{\cos^2 x}} + \frac{\frac{\cos^3 x}{\sin^3 x}}{\frac{1}{\sin^2 x}}$
$= \frac{\sin^3 x}{\cos^3 x} \times \frac{\cos^2 x}{1} + \frac{\cos^3 x}{\sin^3 x} \times \frac{\sin^2 x}{1}$
$= \frac{\sin^3 x}{\cos x} + \frac{\cos^3 x}{\sin x}$
$= \frac{\sin^4 x + \cos^4 x}{\sin x \cos x}$
$= \frac{\left( \sin^2 x \right)^2 + \left( \cos^2 x \right)^2}{\sin x \cos x}$
$= \frac{\left( \sin^2 x + \cos^2 x \right)^2 - 2 \sin^2 x \cos^2 x}{\sin x \cos x}$
$= \frac{1^2 - 2 \sin^2 x \cos^2 x}{\sin x \cos x}$
$= \frac{1 - 2 \sin^2 x \cos^2 x}{\sin x \cos x}$
= RHS
Hence proved.

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 5 Trigonometric Functions
Exercise 5.1 | Q 10 | Page 18