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प्रश्न
Prove the following identities
\[\left( \sec x \sec y + \tan x \tan y \right)^2 - \left( \sec x \tan y + \tan x \sec y \right)^2 = 1\]
उत्तर
\[\text{ LHS }= \left( \sec x \sec y + \tan x \tan y \right)^2 - \left( \sec x \tan y + \tan x \sec y \right)^2 \]
\[ = \left[ \left( \sec x \sec y \right)^2 + \left( \tan x \tan y \right)^2 - 2\left( \sec x \sec y \right)\left( \tan x \tan y \right) \right] \]
\[ - \left[ \left( \sec x \tan y \right)^2 + \left( \tan x \sec y \right)^2 - 2\left( \sec x \tan y \right)\left( \tan x \sec y \right) \right] \]
\[ = \left[ \sec^2 x \sec^2 y + \tan^2 x \tan^2 y - 2\sec x \sec y \tan x \tan y \right] \]
\[ - \left[ \sec^2 x \tan^2 y + \tan^2 x \sec^2 y - 2 \sec x \sec y \tan x \tan y \right]\]
\[ = \sec^2 x \sec^2 y + \tan^2 x \tan^2 y - 2 \sec x \sec y \tan x \tan y \]
\[ - \sec^2 x \tan^2 y - \tan^2 x \sec^2 y + 2 \sec x \sec y \tan x \tan y\]
\[ = \sec^2 x \sec^2 y - \sec^2 x \tan^2 y + \tan^2 x \tan^2 y - \tan^2 x \sec^2 y\]
\[ = \sec^2 x \left( \sec^2 y - \tan^2 y \right) + \tan^2 x\left( \tan^2 y - \sec^2 y \right)\]
\[ = \sec^2 x \left( \sec^2 y - \tan^2 y \right) - \tan^2 x\left( \sec^2 y - \tan^2 y \right)\]
\[ = \sec^2 x \times 1 - \tan^2 x \times 1\]
\[ = \sec^2 x - \tan^2 x\]
\[ = 1\]
= RHS
Hence proved.
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