Question

If $\sin \theta =\dfrac{a}{b}$, then find the value of $\sec \theta +\tan \theta$ in terms of a and b.

Answer 1

Hint:First of all we will use $si{{n}^{2}}\theta +{{\cos }^{2}}\theta =1$ to find the value of $\cos \theta$. Now use $sec\theta =\dfrac{1}{\cos \theta }$ to find the value of $\sec \theta$. Now use $1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta$ to find the value of $\tan \theta$. Now substitute these values in the expression $\sec \theta +\tan \theta$ to find the required answer.

Complete step-by-step answer:
Here, we are given that $\sin \theta =\dfrac{a}{b}$. We have to find the value of $\sec \theta +\tan \theta$ in terms of a and b.
Let us consider the expression asked in question.
$E=\sec \theta +\tan \theta .....(1)$
We are given that $\sin \theta =\dfrac{a}{b}$
We know that $si{{n}^{2}}\theta +{{\cos }^{2}}\theta =1$
By substituting the values of $\sin \theta$, we get as follows:

& {{\left( \dfrac{a}{b} \right)}^{2}}+{{\cos }^{2}}\theta =1 \\\ & {{\cos }^{2}}\theta =1-\dfrac{{{a}^{2}}}{{{b}^{2}}} \\\ & \Rightarrow \cos \theta =\sqrt{\dfrac{{{b}^{2}}-{{a}^{2}}}{{{b}^{2}}}}=\dfrac{\sqrt{{{b}^{2}}-{{a}^{2}}}}{b} \\\ \end{aligned}$$ We also know that $$\sec \theta =\dfrac{1}{\cos \theta }$$ So by substituting the value of $$\cos \theta $$, we get as follows: $$\sec \theta =\dfrac{1}{\dfrac{\sqrt{{{b}^{2}}-{{a}^{2}}}}{b}}=\dfrac{b}{\sqrt{{{b}^{2}}-{{a}^{2}}}}......(2)$$ Now we know that $$1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta $$. So by substituting the value of $$\sec \theta $$ we get as follows: $$\begin{aligned} & 1+{{\tan }^{2}}\theta ={{\left( \dfrac{b}{\sqrt{{{b}^{2}}-{{a}^{2}}}} \right)}^{2}} \\\ & 1+{{\tan }^{2}}\theta =\dfrac{{{b}^{2}}}{{{b}^{2}}-{{a}^{2}}} \\\ & {{\tan }^{2}}\theta =\dfrac{{{b}^{2}}}{{{b}^{2}}-{{a}^{2}}}-1 \\\ & {{\tan }^{2}}\theta =\dfrac{{{b}^{2}}-\left( {{b}^{2}}-{{a}^{2}} \right)}{{{b}^{2}}-{{a}^{2}}} \\\ & {{\tan }^{2}}\theta =\dfrac{{{b}^{2}}-{{b}^{2}}+{{a}^{2}}}{{{b}^{2}}-{{a}^{2}}} \\\ & {{\tan }^{2}}\theta =\dfrac{{{a}^{2}}}{{{b}^{2}}-{{a}^{2}}} \\\ & \tan \theta =\sqrt{\dfrac{{{a}^{2}}}{{{b}^{2}}-{{a}^{2}}}} \\\ & \tan \theta =\dfrac{a}{\sqrt{{{b}^{2}}-{{a}^{2}}}} \\\ \end{aligned}$$ So we get $$\tan \theta =\dfrac{a}{\sqrt{{{b}^{2}}-{{a}^{2}}}}......(3)$$ Now by substituting the values of $$\sec \theta $$ and $$\tan \theta $$ from equation (2) and (3) in equation (1), we get as follows: $$\begin{aligned} & E=\sec \theta +\tan \theta \\\ & E=\dfrac{b}{\sqrt{{{b}^{2}}-{{a}^{2}}}}+\dfrac{a}{\sqrt{{{b}^{2}}-{{a}^{2}}}} \\\ & E=\dfrac{(b+a)}{\sqrt{{{b}^{2}}-{{a}^{2}}}} \\\ \end{aligned}$$ So we get the value of $$\sec \theta +\tan \theta $$ as $$\dfrac{(b+a)}{\sqrt{{{b}^{2}}-{{a}^{2}}}}$$. Note: In this type of questions, we can also find various trigonometric ratios by considering right angled triangle ABC and angle C as $$\theta $$. Now take sides AB and AC as a and b respectively and find BC by Pythagoras theorem. Now find $$\sec \theta $$ by using $$\sec \theta =\dfrac{hypotenuse}{base}$$ and $$\tan \theta $$ by using $$\tan \theta =\dfrac{perpendicular}{base}$$.Substitute the values in the given expression to find the answer.