MCQ
If $x = \sec \theta + \tan \theta ,$ then $x + \frac{1}{x} = $
- A$1$
- ✓$2\sec \theta $
- C$2$
- D$2\tan \theta $
$ \Rightarrow \,x + \frac{1}{x} = \sec \theta + \tan \theta + \frac{1}{{\sec \theta + \tan \theta }}$
$ = \sec \theta + \tan \theta + \sec \theta - \tan \theta = 2\sec \theta $
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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.