Find the values of other five trigonometric functions if tan x = -5/12

Question:

Find the values of other five trigonometric functions if $\tan x=-\frac{5}{12}, x$ lies in second quadrant.

Solution:

$\tan x=-\frac{5}{12}$

$\cot x=\frac{1}{\tan x}=\frac{1}{\left(-\frac{5}{12}\right)}=-\frac{12}{5}$

$1+\tan ^{2} x=\sec ^{2} x$

$\Rightarrow 1+\left(-\frac{5}{12}\right)^{2}=\sec ^{2} x$

$\Rightarrow 1+\frac{25}{144}=\sec ^{2} x$

$\Rightarrow \frac{169}{144}=\sec ^{2} x$

$\Rightarrow \sec x=\pm \frac{13}{12}$

Since $x$ lies in the $2^{\text {nd }}$ quadrant, the value of $\sec x$ will be negative.

$\therefore \sec x=-\frac{13}{12}$

$\cos x=\frac{1}{\sec x}=\frac{1}{\left(-\frac{13}{12}\right)}=-\frac{12}{13}$

$\tan x=\frac{\sin x}{\cos x}$

$\Rightarrow-\frac{5}{12}=\frac{\sin x}{\left(-\frac{12}{13}\right)}$

$\Rightarrow \sin x=\left(-\frac{5}{12}\right) \times\left(-\frac{12}{13}\right)=\frac{5}{13}$

$\operatorname{cosec} x=\frac{1}{\sin x}=\frac{1}{\left(\frac{5}{13}\right)}=\frac{13}{5}$

 

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now