MCQ
If $\tan \left( {\frac{\pi }{4} + \theta } \right) + \tan \left( {\frac{\pi }{4} - \theta } \right) = \lambda \sec 2\theta ,$ $\lambda$ =
  • A
    $3$
  • B
    $4$
  • C
    $1$
  • $2$

Answer

Correct option: D.
$2$
d
$\frac{{1 + \tan \theta }}{{1 - \tan \theta }} + \frac{{1 - \tan \theta }}{{1 + \tan \theta }} = \lambda \left[ {\frac{{1 + {{\tan }^2}\theta }}{{1 - {{\tan }^2}\theta }}} \right]$

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