Question

If $\cot A=\frac{12}{5}$, then find the value of $(\sin A+\cos A)\times cosecA$.

• A. $\frac{13}{5}$
• B. $\frac{17}{5}$
• C. $\frac{14}{5}$
• D. 1

Answer 1

The correct option is B

$\frac{17}{5}$

$\cot A=\frac{Base}{Perpendicular}=\frac{12}{5}$

So, if base = 12k, then perpendicular = 5k, where k is a positive number.

$Hypotenuse=\sqrt{bas{e}^2+perpendicula{r}^2}$ (using Pythagoras theorem)

= $\sqrt{(12k{)}^2+(5k{)}^2}$
= 13k

Now, $\sin A=\frac{5k}{13k}=\frac{5}{13},\cos A=\frac{12k}{13k}=\frac{12}{13}$

and $cosecA=\frac{13k}{5k}=\frac{13}{5}$

So, $(\sin A+\cos A)\times cosecA$

$=(\frac{5}{13}+\frac{12}{13})\times \frac{13}{5}=\frac{17}{5}$