Question

$3\tan x+4=0$ and $\frac{\pi }{2}<\theta <\pi$ then $2\cot \theta -5\cos \theta +\sin \theta =.............$

• A. $\frac{56}{10}$
• B. $\frac{23}{10}$
• C. $\frac{37}{10}$
• D. $\frac{7}{10}$

Answer 1

Answer: (B)

$\frac{23}{10}$

Given : $3\tan x+4=0\Rightarrow \tan x=-\frac{4}{3}$

Now $\frac{\pi }{2}<x<\pi$

$\therefore \cot x=-\frac{3}{4}$ $\cos x=-\frac{3}{5}$ $\sin \theta =\frac{4}{5}$

$\therefore 2\cot x-5\cos x+\sin x$

$-2\times \frac{3}{4}-5(-\frac{3}{5})+\frac{4}{5}$

$-\frac{6}{4}+3+\frac{4}{5}=\frac{-30+60+16}{20}=\frac{46}{20}=\frac{23}{10}$