Question

If $\theta =\frac{11\pi }{4}$, find the value of ${\sin }^2\theta -{\cos }^2\theta +2\tan \theta -{\sec }^2\theta$

Answer 1

$\theta =\frac{11\pi }{4}=(3\pi -\frac{\pi }{4})$
$\sin \theta =\sin (3\pi -\frac{\pi }{4})=\sin \frac{\pi }{4}=\sin {45}^0=\frac{1}{\sqrt{2}}$
$\cos \theta =\cos (3\pi -\frac{\pi }{4})=-\cos \frac{\pi }{4}$
$=-\cos {45}^0=-\frac{1}{\sqrt{2}}$
$\tan \theta =\tan (3\pi -\frac{\pi }{4})=-\tan \frac{\pi }{4}=-\tan {45}^0=-1$
$\therefore$ Given expression $=\frac{1}{2}-\frac{1}{2}-2-2=-4$