MCQ
Let $0 < x < \frac{\pi }{4}.$ Then $\sec 2x - \tan 2x = $
- A$\tan \left( {x - \frac{\pi }{4}} \right)$
- ✓$\tan \left( {\frac{\pi }{4} - x} \right)$
- C$\tan \left( {x + \frac{\pi }{4}} \right)$
- D${\tan ^2}\left( {x + \frac{\pi }{4}} \right)$
$ = \frac{{{{(\cos x - \sin x)}^2}}}{{({{\cos }^2}x - {{\sin }^2}x)}} $
$= \frac{{\cos x - \sin x}}{{\cos x + \sin x}} = \frac{{1 - \tan x}}{{1 + \tan x}}$
$ = \frac{{\tan \frac{\pi }{4} - \tan x}}{{1 + \tan \left( {\frac{\pi }{4}} \right)\sin x}} = \tan \left( {\frac{\pi }{4} - x} \right)$.
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Statement $p$ : The value of $sin\,120^o$ can be divided by taking $\theta\, = 240^o$ in the equation $2\,\sin \frac{\theta }{2} = \sqrt {1 + \sin \theta } - \sqrt {1 - \sin \theta } $
Statement $q$ : The angles $A, B, C$ and $D$ of any quadrilateral $ABCD$ satisfy the equation $\cos \left( {\frac{1}{2}\left( {A + C} \right)} \right) + \cos \left( {\frac{1}{2}\left( {B + D} \right)} \right) = 0$
Then the truth values of $p$ and $q$ are respectively.