$1 - 2{\sin ^2}\left( {\frac{\pi }{4} + \theta } \right) = $
$1 - 2{\sin ^2}\left( {\frac{\pi }{4} + \theta } \right) = $
- A
$\cos 2\theta $
- B
$ - \cos 2\theta $
- C
$\sin 2\theta $
- D
$ - \sin 2\theta $
Similar Questions
$(\sec 2A + 1){\sec ^2}A = $
$(\sec 2A + 1){\sec ^2}A = $
The value of $cot\, 7\frac{{{1^0}}}{2}$ $+ tan\, 67 \frac{{{1^0}}}{2} - cot 67 \frac{{{1^0}}}{2} - tan7 \frac{{{1^0}}}{2}$ is :
The value of $cot\, 7\frac{{{1^0}}}{2}$ $+ tan\, 67 \frac{{{1^0}}}{2} - cot 67 \frac{{{1^0}}}{2} - tan7 \frac{{{1^0}}}{2}$ is :
$\sqrt 2 + \sqrt 3 + \sqrt 4 + \sqrt 6 $ is equal to
$\sqrt 2 + \sqrt 3 + \sqrt 4 + \sqrt 6 $ is equal to
- [IIT 1966]
The value of $x$ that satisfies the relation $x = 1 - x + x^2 - x^3 + x^4 - x^5 + ......... \infty$
The value of $x$ that satisfies the relation $x = 1 - x + x^2 - x^3 + x^4 - x^5 + ......... \infty$
$\left( {\frac{{\sin 2A}}{{1 + \cos 2A}}} \right)\,\left( {\frac{{\cos A}}{{1 + \cos A}}} \right)= $
$\left( {\frac{{\sin 2A}}{{1 + \cos 2A}}} \right)\,\left( {\frac{{\cos A}}{{1 + \cos A}}} \right)= $