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Prove the Following Identities ( 1 + Tan α Tan β ) 2 + ( Tan α − Tan β ) 2 = Sec 2 α Sec 2 β - Mathematics

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Question

Prove the following identities
\[\left( 1 + \tan \alpha \tan \beta \right)^2 + \left( \tan \alpha - \tan \beta \right)^2 = \sec^2 \alpha \sec^2 \beta\]

Solution

\[\left( 1 + \tan \alpha \tan \beta \right)^2 + \left( \tan \alpha - \tan \beta \right)^2 = \sec^2 \alpha \sec^2 \beta\]

\[\text{ LHS }= \left( 1 + \tan\alpha \tan\beta \right)^2 + \left( \tan\alpha - \tan\beta \right)^2 \]

\[ = 1 + \tan^2 \alpha \tan^2 \beta + 2\tan\alpha \tan\beta + \tan^2 \alpha + \tan^2 \beta - 2\tan\alpha \tan\beta\]

\[ = 1 + \tan^2 \alpha \tan^2 \beta + \tan^2 \alpha + \tan^2 \beta\]

\[ = \tan^2 \alpha\left( \tan^2 \beta + 1 \right) + 1\left( 1 + \tan^2 \beta \right)\]

\[ = \left( 1 + \tan^2 \beta \right)\left( 1 + \tan^2 \alpha \right)\]

\[ = \sec^2 \alpha . \sec^2 \beta \]

= RHS

Hence proved.

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Trigonometric Functions - Truth of the Identity
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Chapter 5: Trigonometric Functions - Exercise 5.1 [Page 18]

APPEARS IN

RD Sharma Mathematics [English] Class 11
Chapter 5 Trigonometric Functions
Exercise 5.1 | Q 13 | Page 18

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