Question

If $A$ lies in the second quadrant and $3\tan A+4=0$ then the value of $2cotA-5\cos A+\sin A$ is equal to

• A. $\frac{-53}{10}$
• B. $\frac{-7}{10}$
• C. $\frac{7}{10}$
• D. $\frac{23}{10}$

Answer 1

The correct option is D

$\frac{23}{10}$

Explanation for the correct option:

Step 1. Find the value of $2cotA-5\cos A+\sin A$:

Given, $A$ lies in the second quadrant.

$3\tan A+4=0$

$\Rightarrow$ $\tan A=\frac{-4}{3}$

$\Rightarrow$ $cotA=\frac{-3}{4}$

Step 2. Find $\cos A$ and $\sin A$:

$\begin{array}{rcl}se{c}^{2}A& =& 1+{\tan }^{2}A\\ & =& 1+\frac{16}{9}\\ & =& \frac{25}{9}\end{array}$

$secA=\sqrt{\frac{25}{9}}$

$=\frac{-5}{3}$ [$\because$in quadrant II secant is negative]

$\cos A=\frac{-3}{5}$

$\begin{array}{rcl}\sin A& =& \sqrt{1–\frac{9}{25}}\\ & =& \sqrt{\frac{16}{25}}\end{array}$

$\begin{array}{rcl}& =& \frac{4}{5}\end{array}$ [$\because$in quadrant II sine is positive]

Step 3. Put the values of $\cos A$ and $\sin A$ in given expression:

$\begin{array}{rcl}\therefore 2cotA–5\cos A+\sin A& =& 2\left(\frac{-3}{4}\right)–5\left(\frac{-3}{5}\right)+\left(\frac{4}{5}\right)\\ & =& \left(\frac{-3}{2}\right)+3+\left(\frac{4}{5}\right)\\ & =& \frac{(-15+30+8)}{10}\\ & =& \frac{\mathbf{23}}{\mathbf{10}}\end{array}$

Hence, Option ‘D’ is Correct.