Question

If $\cot A+\text{cosec}A=3$ and $A$ is an acute angle, then the value of $\cos A$ is

• A. $\frac{4}{5}$
• B. $\frac{3}{5}$
• C. $\frac{1}{2}$
• D. $\frac{1}{5}$

Answer 1

The correct option is A $\frac{4}{5}$

$\cot A+\text{cosec}A=3...\left(1\right)$
We know that,
${\text{cosec}}^{2}A-{\cot }^{2}A=1$
$\Rightarrow (\text{cosec}A+\cot A)(\text{cosec}A-\cot A)=1$
$\Rightarrow \text{cosec}A-\cot A=\frac{1}{3}...\left(2\right)$
From $\left(1\right)$ and $\left(2\right),$
$\text{cosec}A=\frac{5}{3}$
Base $=4$
$\therefore \cos A=\frac{4}{5}$


Alternate Solution:
$\Rightarrow \frac{\cos A}{\sin A}+\frac{1}{\sin A}=3$
$\Rightarrow \cos A+1=3\sin A$
squaring both sides
$\Rightarrow (\cos A+1{)}^2=(3\sin A{)}^2$
$\Rightarrow {\cos }^{2}A+2\cos A+1=9{\sin }^{2}A$
$\Rightarrow {\cos }^{2}A+2\cos A+1=9(1-{\cos }^{2}A)$
$\Rightarrow 10{\cos }^{2}A+2\cos A-8=0$
$\Rightarrow 5{\cos }^{2}A+\cos A-4=0$
$\Rightarrow (5\cos A-4)(\cos A+1)=0$
$\Rightarrow \cos A=\frac{4}{5},-1$
Since, $A$ is an acute angle
$\therefore \cos A$ cannot be negative.
$\therefore \cos A=\frac{4}{5}$