# If Cot X ( 1 + Sin X ) = 4 M and Cot X ( 1 − Sin X ) = 4 N , ( M 2 + N 2 ) 2 = M N - Mathematics

If $\cot x \left( 1 + \sin x \right) = 4 m \text{ and }\cot x \left( 1 - \sin x \right) = 4 n,$ $\left( m^2 + n^2 \right)^2 = mn$

#### Solution

Given:

$4m = cotx\left( 1 + \sin x \right) and 4n = cot x\left( 1 - \sin x \right)$

Multiplying both the equations:

$\Rightarrow 16mn = co t^2 x\left( 1 - \sin^2 x \right)$

$\Rightarrow 16mn = co t^2 x . \cos^2 x$

$\Rightarrow mn = \frac{\cos^4 x}{16 \sin^2 x} \left( 1 \right)$

Squaring the given equation:

$16 m^2 = co t^2 x \left( 1 + \sin x \right)^2 \text{ and }16 n^2 = co t^2 x \left( 1 - \sin x \right)^2$

$\Rightarrow 16 m^2 - 16 n^2 = co t^2 x\left( 4\sin x \right)$

$\Rightarrow m^2 - n^2 = \frac{co t^2 x . \sin x}{4}$

Squaring both sides,

$\left( m^2 - n^2 \right)^2 = \frac{co t^4 x . \sin^2 x}{16}$

$\Rightarrow \left( m^2 - n^2 \right)^2 = \frac{\cos^4 x}{16 \sin^2 x} (2)$

From (1) and (2):

$\left( m^2 - n^2 \right)^2 = mn$

Hence proved.

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 5 Trigonometric Functions
Exercise 5.1 | Q 22 | Page 19