If $\tan \theta - \cot \theta = a$ and $\sin \theta + \cos \theta = b,$ then ${({b^2} - 1)^2}({a^2} + 4)$ is equal to

  • A

    $2$

  • B

    $-4$

  • C

    $± 4$

  • D

    $4$

Similar Questions

$\cos 15^\circ = $

$(m + 2)\sin \theta + (2m - 1)\cos \theta = 2m + 1,$ if

Prove that $\cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)-\sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)=\sin (x+y)$

If $\cos \theta - \sin \theta = \sqrt 2 \sin \theta ,$ then $\cos \theta + \sin \theta $ is equal to

If $x\sin 45^\circ {\cos ^2}60^\circ = \frac{{{{\tan }^2}60^\circ {\rm{cosec}}30^\circ }}{{\sec 45^\circ {{\cot }^2}30^\circ }},$ then $x = $