Question

Given that $3\sin \theta +4\cos \theta =5where\theta \in \left(0,\frac{\pi }{2}\right).$ Find the value of $2\sin \theta +\cos \theta +4\tan \theta +3\cot \theta$

Answer 1

$3\sin \theta +4\cos \theta =5$

$\frac{3}{5}\sin \theta +\frac{4}{3}\cos \theta =1$

Let $\cos t=\frac{3}{5}$

So, $\sin t=\sqrt{1-{\cos }^{2}t}$

$=\sqrt{1-\frac{a}{2s}}$

$=\sqrt{\frac{16}{25}}=\frac{4}{3}$

$\sin \theta \cos t+\cos \theta \sin t=1$

$\sin (\theta +t)=1$

$\theta +t=\frac{\pi }{2}$

$\theta =\frac{\pi }{2}-t$

So, $2\sin \theta +\cos \theta +4\tan \theta +3\cot \theta$

$=2\sin (\frac{\pi }{2}-t)+\cos (\frac{\pi }{2}-t)+4(\cos (\frac{\pi }{2}-t)+3\cot (\frac{\pi }{2}-t))$

$=2\cos t+\sin t+4\cot t+3\tan t$

$=2\times \frac{3}{5}+\frac{4}{5}+4\frac{\cos t}{\sin t}+\frac{3\sin t}{\cos t}$

$=\frac{6}{5}+\frac{4}{5}+4\times \frac{3}{5}\times \frac{5}{5}+3\times \frac{4}{3}\times \frac{5}{3}$

$=\frac{10}{5}+3+4.2+7=9$ (Answer)$