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Question
If \[\tan \left( \pi/4 + x \right) + \tan \left( \pi/4 - x \right) = \lambda \sec 2x, \text{ then } \]
Options
3
4
1
2
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Solution
2
\[\text{ Given } : \]
\[ \tan\left( \frac{\pi}{4} + x \right) + \tan\left( \frac{\pi}{4} - x \right) = \lambda \sec 2x\]
\[ \Rightarrow \frac{\tan\frac{\pi}{4} + \text{ tan } x}{1 - \tan\frac{\pi}{4} \times \text{ tan } x} + \frac{\tan\frac{\pi}{4} - \text{ tan } x}{1 + \tan\frac{\pi}{4} \times \text{ tan } x} = \lambda \sec 2x\]
\[ \Rightarrow \frac{1 + \text{ tan } x}{1 - \text{ tan } x} + \frac{1 - \text{ tan } x}{1 + \text{ tan } x} = \lambda \sec 2x\]
\[ \Rightarrow \frac{\left( 1 + \text{ tan } x \right)^2 + \left( 1 - \text{ tan } x \right)^2}{\left( 1 - \text{ tan } x \right)\left( 1 + \text{ tan } x \right)} = \lambda \sec 2x\]
\[ \Rightarrow \frac{2\left( 1 + \tan^2 x \right)}{1 - \tan^2 x} = \lambda \sec 2x\]
\[\Rightarrow \frac{2 \sec^2 x}{1 - \tan^2 x} = \lambda \sec 2x\]
\[ \Rightarrow \frac{2}{\cos^2 x\left( 1 - \tan^2 x \right)} = \lambda \sec 2x\]
\[ \Rightarrow \frac{2}{\cos^2 x\left( 1 - \frac{\sin^2 x}{\cos^2 x} \right)} = \lambda \sec 2x\]
\[ \Rightarrow \frac{2}{\cos^2 x - \sin^2 x} = \lambda \sec 2x\]
\[ \Rightarrow \frac{2}{\cos2x} = \lambda \sec 2x\]
\[ \Rightarrow 2\sec2x = \lambda \sec 2x\]
\[ \Rightarrow 2 = \lambda\]
\[ \therefore \lambda = 2\]
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