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 $\cos \theta-4 \sin \theta=1 \ldots$ (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 $\theta=\pi$
$\sin \theta+4 \cos \theta=0-4=-4$