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

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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
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