Question

If $\sin \theta =\cos (2\theta -{45}^{\circ }),0<(2\theta -{45}^{\circ })<{90}^{\circ },$ then $\tan \theta$

• A. $-1$
• B. $0$
• C. $1$
• D. $\frac{1}{\sqrt{3}}$

Answer 1

Answer: (C)

$1$

Given, $\sin \theta =\cos (2\theta -{45}^{\circ })$
since, $\sin \theta =\cos ({90}^{\circ }-\theta )$
$\therefore 2\theta -{45}^{\circ }={90}^{\circ }-\theta$
$\therefore 3\theta ={135}^{\circ }$
$\therefore \theta ={45}^{\circ }$
So, $\tan \theta =\tan {45}^{\circ }=1$