Let $0 < x < \frac{\pi }{4}.$ Then $\sec 2x - \tan 2x = $
Let $0 < x < \frac{\pi }{4}.$ Then $\sec 2x - \tan 2x = $
- [IIT 1994]
- A
$\tan \left( {x - \frac{\pi }{4}} \right)$
- B
$\tan \left( {\frac{\pi }{4} - x} \right)$
- C
$\tan \left( {x + \frac{\pi }{4}} \right)$
- D
${\tan ^2}\left( {x + \frac{\pi }{4}} \right)$
Similar Questions
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The value of $2 \sin(\frac{\pi}{8}) \sin (\frac{2 \pi}{8}) \sin (\frac{3 \pi}{8}) \sin (\frac{5 \pi}{8}) \sin (\frac{6 \pi}{8}) \sin (\frac{7 \pi}{8})$ is:
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- [JEE MAIN 2021]
If $\alpha$, $\beta$,$\gamma$ are positive number such that $\alpha + \beta = \pi$ and $\beta + \gamma = \alpha$, then $tan\ \alpha$ is equal to - (where $\gamma \ne n\pi ,n \in I$ )
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$cosec^2\theta $ = $\frac{4xy}{(x +y)^2}$ is true if and only if
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