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:
\[\begin{align}
  & {{\left( \dfrac{a}{b} \right)}^{2}}+{{\cos }^{2}}\theta =1 \\
 & {{\cos }^{2}}\theta =1-\dfrac{{{a}^{2}}}{{{b}^{2}}} \\
 & \Rightarrow \cos \theta =\sqrt{\dfrac{{{b}^{2}}-{{a}^{2}}}{{{b}^{2}}}}=\dfrac{\sqrt{{{b}^{2}}-{{a}^{2}}}}{b} \\
\end{align}\]
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:
\[\begin{align}
  & 1+{{\tan }^{2}}\theta ={{\left( \dfrac{b}{\sqrt{{{b}^{2}}-{{a}^{2}}}} \right)}^{2}} \\
 & 1+{{\tan }^{2}}\theta =\dfrac{{{b}^{2}}}{{{b}^{2}}-{{a}^{2}}} \\
 & {{\tan }^{2}}\theta =\dfrac{{{b}^{2}}}{{{b}^{2}}-{{a}^{2}}}-1 \\
 & {{\tan }^{2}}\theta =\dfrac{{{b}^{2}}-\left( {{b}^{2}}-{{a}^{2}} \right)}{{{b}^{2}}-{{a}^{2}}} \\
 & {{\tan }^{2}}\theta =\dfrac{{{b}^{2}}-{{b}^{2}}+{{a}^{2}}}{{{b}^{2}}-{{a}^{2}}} \\
 & {{\tan }^{2}}\theta =\dfrac{{{a}^{2}}}{{{b}^{2}}-{{a}^{2}}} \\
 & \tan \theta =\sqrt{\dfrac{{{a}^{2}}}{{{b}^{2}}-{{a}^{2}}}} \\
 & \tan \theta =\dfrac{a}{\sqrt{{{b}^{2}}-{{a}^{2}}}} \\
\end{align}\]
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:
\[\begin{align}
  & E=\sec \theta +\tan \theta \\
 & E=\dfrac{b}{\sqrt{{{b}^{2}}-{{a}^{2}}}}+\dfrac{a}{\sqrt{{{b}^{2}}-{{a}^{2}}}} \\
 & E=\dfrac{(b+a)}{\sqrt{{{b}^{2}}-{{a}^{2}}}} \\
\end{align}\]
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.