Question

If sin \[x = \frac{12}{13}\] and x lies in the second quadrant, find the value of sec x + tan x.

Answer 1

We have:
\[\sin x = \frac{12}{13}\text{ and x lie in the second quadrant . }\]
In the second quadrant, sin x and cosec x are positive and all the other four T - ratios are negative .
\[\therefore \cos x = - \sqrt{1 - \sin^2 x}\]
\[ = - \sqrt{1 - \left( \frac{12}{13} \right)^2} \]
\[ = \frac{- 5}{13}\]
\[\tan x = \frac{\sin x}{\cos x} \]
\[ = \frac{\frac{12}{13}}{\frac{- 5}{13}}\]
\[ = \frac{- 12}{5}\]
\[\text{ And, }\sec x = \frac{1}{\cos x} \]
\[ = \frac{1}{\frac{- 5}{13}} \]
\[ = \frac{- 13}{5}\]
\[ \therefore \sec x + \tan x = \frac{- 13}{5} + \frac{- 12}{5}\]
\[ = - 5\]