Question

If A lies in second quadrant 3tanA + 4 = 0, then the value of 2cotA − 5cosA + sinA is equal to

(a) $-\frac{53}{10}$ (b) $\frac{23}{10}$ (c) $\frac{37}{10}$ (d) $\frac{7}{10}$

Answer 1

It is given that $\frac{\mathrm{\pi }}{2}<A<\mathrm{\pi }$.

$$\begin{gathered} 3\tan A+4=0 \\ \Rightarrow \tan A=-\frac{4}{3} \\ \Rightarrow \cot A=-\frac{3}{4} \end{gathered}$$

Now,

$$\begin{gathered} \sec A=\pm \sqrt{1+{\tan }^{2}A}=\pm \sqrt{1+\frac{16}{9}}=\pm \sqrt{\frac{25}{9}}=\pm \frac{5}{3} \\ \therefore \sec A=-\frac{5}{3}\left(A\mathrm{lies}\mathrm{in}2\mathrm{nd}\mathrm{quadrant}\right) \\ \Rightarrow \cos A=-\frac{3}{5} \end{gathered}$$

Also,

$$\begin{gathered} \sin A=\pm \sqrt{1-{\cos }^{2}A}=\pm \sqrt{1-\frac{9}{25}}=\pm \sqrt{\frac{16}{25}}=\pm \frac{4}{5} \\ \therefore \sin A=\frac{4}{5}\left(A\mathrm{lies}\mathrm{in}2\mathrm{nd}\mathrm{quadrant}\right) \end{gathered}$$

So,

$$\begin{gathered} 2\cot A-5\cos A+\sin A \\ =2\times \left(-\frac{3}{4}\right)-5\times \left(-\frac{3}{5}\right)+\frac{4}{5} \\ =-\frac{3}{2}+3+\frac{4}{5} \\ =\frac{-15+30+8}{10} \\ =\frac{23}{10} \end{gathered}$$

Hence, the correct answer is option B.