Question

If $\cos \theta -4\sin \theta =1$, then $\sin \theta +4\cos \theta$ is equal to

• A. $\pm 1$
• B. 0
• C. $\pm 2$
• D. $\pm 4$

Answer 1

Answer: (D)

Given that $\cos \theta -4\sin \theta =1\dots$ (i)
On squaring both sides of equation (i)
${\cos }^2\theta +16{\sin }^2\theta -8\sin \theta \cos \theta =1$
$\Rightarrow 15{\sin }^2\theta -8\sin \theta \cos \theta =0$
$\Rightarrow \sin \theta (15\sin \theta -8\cos \theta )=0$
$\Rightarrow \sin \theta =0$ or $\tan \theta =\frac{8}{15}$
but $\tan \theta$ is not satisfy the equation (i)
$\therefore \sin \theta =0$
$\Rightarrow \theta =0,\pi$
at $\theta =0$
$\sin \theta +4\cos \theta =0+4=4$
at $\theta =\pi$
$\sin \theta +4\cos \theta =0-4=-4$