Question

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

Answer 1

We have,

$si{n}^2\theta +co{s}^2\theta =1$

$\Rightarrow co{s}^2\theta =1-si{n}^2\theta$

$\Rightarrow cos\theta =\pm \sqrt{1-si{n}^2\theta }$

In the 2st quadrant cos $\theta$ is negative and tan $\theta$ is also negative

$\therefore cos\theta =-\sqrt{1-si{n}^2\theta }$

$=-\sqrt{1-(\frac{12}{13}{)}^2}[\because sin\theta =\frac{12}{13}]$

$=-\sqrt{1-\frac{144}{169}}$

$=-\sqrt{\frac{25}{169}}=-\frac{5}{13}$

and, $tan\theta =\frac{sin\theta }{cos\theta }=\frac{\frac{12}{13}}{\frac{-5}{13}}=-\frac{13}{5}$

Now, $sec\theta =\frac{1}{cos\theta }=\frac{1}{-5}=-\frac{13}{5}$

$\therefore sec\theta +tan\theta =-\frac{13}{5}-\frac{12}{5}$

$=\frac{-13-12}{5}$

$=-\frac{25}{5}$

=-5

$\Rightarrow sec\theta +tan\theta =-5$