Question

$cos\theta -4sin\theta =1\Rightarrow sin\theta +4cos\theta$

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

Answer 1

The correct option is D $\pm 4$

Let $\sin \theta +4\cos \theta =k\cdots \left(1\right)$

Given $\cos \theta -4\sin \theta =1\cdots \left(2\right)$

$(1{)}^2+(2{)}^2⟹(\sin \theta +4\cos \theta {)}^2+(\cos \theta -4\sin \theta {)}^2=k^2+1^2$

$⟹{\sin }^2\theta +16{\cos }^2\theta +8\sin \theta \cos \theta +{\cos }^2\theta +16{\sin }^2\theta -8\sin \theta \cos \theta =k^2+1$

$⟹({\sin }^2\theta +{\cos }^2\theta )+16({\sin }^2\theta +{\cos }^2\theta )=k^2+1$

$⟹k^2=1+16-1=16⟹k=\pm 4$

So $\sin \theta +4\cos \theta =\pm 4$