Question

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

Answer 1

Hint:First of all, try to recollect the sign of $\sec \theta$ and $\tan \theta$ in the second quadrant. Now, first find $\cos \theta$ by using $\sqrt{1-{{\sin }^{2}}\theta }$ and taking its reciprocal to find $\sec \theta$. Use $\dfrac{\sin \theta }{\cos \theta }$ to find $\tan \theta$. Finally, add these values to get $\sec \theta +\tan \theta$.

Complete step-by-step answer:
We are given that $\sin \theta =\dfrac{12}{13}$ and $\theta$ lies in the second quadrant. We have to find the value of $\sec \theta +\tan \theta$. Before proceeding with this question, let us see the sign of different trigonometric ratios in different quadrants. We have 6 trigonometric ratios and that are $\sin \theta ,\cos \theta ,\tan \theta ,\cot \theta ,\operatorname{cosec}\theta$ and $\sec \theta$.
1. In the first quadrant, that is from 0 to ${{90}^{o}}$ or 0 to $\dfrac{\pi }{2}$, all the trigonometric ratios are positive.
2. In the second quadrant, that is from ${{90}^{o}}$ to ${{180}^{o}}$ or $\dfrac{\pi }{2}$ to $\pi$, only $\sin \theta$ and $\operatorname{cosec}\theta$ are positive.
3. In the third quadrant, that is from ${{180}^{o}}$ to ${{270}^{o}}$ or $\pi$ to $\dfrac{3\pi }{2}$, only $\tan \theta$ and $\cot \theta$ are positive.
4. In the fourth quadrant, that is from ${{270}^{o}}$ to ${{360}^{o}}$ or $\dfrac{3\pi }{2}$ to $2\pi$, only $\cos \theta$ and $\sec \theta$ are positive.
This cycle would repeat after ${{360}^{o}}$.

In this figure, A means all are positive, S means $\sin \theta$ and $\operatorname{cosec}\theta$ are positive, T means $\tan \theta$ and $\cot \theta$ are positive and C means $\cos \theta$ and $\sec \theta$ are positive.
Here, we are given that $\sin \theta =\dfrac{12}{13}$ and $\theta$ is in the second quadrant. So, we know that in this quadrant only $\sin \theta$ and $\operatorname{cosec}\theta$ are positive. So here, $\sec \theta$ and $\tan \theta$ would be negative.
Now, we know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ or ${{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta$. By substituting $\sin \theta =\dfrac{12}{13}$, we get,
${{\cos }^{2}}\theta =1-{{\left( \dfrac{12}{13} \right)}^{2}}$
${{\cos }^{2}}\theta =1-\dfrac{144}{169}$
${{\cos }^{2}}\theta =\dfrac{169-144}{169}=\dfrac{25}{169}$
$\cos \theta =\sqrt{\dfrac{25}{169}}$
$\cos \theta =\pm \dfrac{5}{13}$
We know that in the second quadrant, $\cos \theta$ is negative. So, $\cos \theta =\dfrac{-5}{13}$.
Now, we know that $\sec \theta =\dfrac{1}{\cos \theta }$.
So, $\sec \theta =\dfrac{1}{\dfrac{-5}{13}}$
$\sec \theta =\dfrac{-13}{5}....\left( i \right)$
We also know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$
So, by substituting the value of $\sin \theta =\dfrac{12}{13}$ and $\cos \theta =\dfrac{-5}{13}$. We get,
$\tan \theta =\dfrac{\left( \dfrac{12}{13} \right)}{\left( \dfrac{-5}{13} \right)}$
$\tan \theta =\left( \dfrac{12}{13} \right).\left( \dfrac{13}{-5} \right)$
By canceling the like terms, we get,
$\tan \theta =\dfrac{-12}{5}....\left( ii \right)$
Now, by adding equation (i) and (ii), we get,
$\sec \theta +\tan \theta =\dfrac{-13}{5}-\dfrac{12}{5}$
$\sec \theta +\tan \theta =\dfrac{-13-12}{5}$
$\sec \theta +\tan \theta =\dfrac{-25}{5}=-5$
So, we have got the value of $\sec \theta +\tan \theta$ as – 5.

Note: In this question, students often take the value of $\sec \theta$ and $\tan \theta$ positive and get the answer as 5 which is wrong. In these types of questions, students should first examine the sign of trigonometric ratios properly and then only proceed to solve the question to get the correct answers.And also students should remember the important trigonometric identities,formulas and standard angles to solve these types of questions.