Question

If $\sin \theta =\dfrac{a}{b}$, show that:
$\sec \theta +\tan \theta =\sqrt{\dfrac{b+a}{b-a}}$

Answer 1

Hint: In this question, from the given values of sin function by using the trigonometric identities we can find the values of the sec and tan functions. Then on substituting the respective values in the given expression of the question we can calculate the left hand side value and the right hand side value.

Complete step-by-step answer:
Then on comparing the values obtained, we get the result.

& {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\\ & sec\theta =\dfrac{1}{\cos \theta } \\\ & \tan \theta =\sin \theta \cdot sec\theta \\\ \end{aligned}$$ Now, from the given question we have $$\sin \theta =\dfrac{a}{b}\ \ \ \ \ ...(a)$$ Now, by using the trigonometric identity which gives the relation between the function that are mentioned in the hint, we get the following substitute the value from the question and as well as from equation (a) $$\begin{aligned} & \Rightarrow {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\\ & \Rightarrow {{\cos }^{2}}\theta =1-{{\left( \dfrac{a}{b} \right)}^{2}} \\\ & \Rightarrow {{\cos }^{2}}\theta =\dfrac{{{b}^{2}}-{{a}^{2}}}{{{b}^{2}}} \\\ & \Rightarrow \cos \theta =\sqrt{\dfrac{{{b}^{2}}-{{a}^{2}}}{{{b}^{2}}}} \\\ \end{aligned}$$ For finding the value of sec function, we could again use the relations given in the hint as follows $$\begin{aligned} & \Rightarrow \left( \dfrac{1}{\cos \theta } \right)=\sec \theta \\\ & \Rightarrow \sec \theta =\left( \sqrt{\dfrac{{{b}^{2}}}{{{b}^{2}}-{{a}^{2}}}} \right)=\left( \dfrac{b}{\sqrt{{{b}^{2}}-{{a}^{2}}}} \right) \\\ \end{aligned}$$ Now, again from the hint, we know that we can get the tan function as follows $$\begin{aligned} & \Rightarrow \tan \theta =\sin \theta \cdot \sec \theta \\\ & \Rightarrow \tan \theta =\dfrac{a}{b}\cdot \left( \sqrt{\dfrac{{{b}^{2}}}{{{b}^{2}}-{{a}^{2}}}} \right) \\\ & \Rightarrow \tan \theta =\dfrac{a}{b}\cdot \left( \dfrac{b}{\sqrt{{{b}^{2}}-{{a}^{2}}}} \right) \\\ & \Rightarrow \tan \theta =\left( \dfrac{a}{\sqrt{{{b}^{2}}-{{a}^{2}}}} \right) \\\ \end{aligned}$$ Now, from the given expression in the question, on substituting the values, we have $$\sec \theta +\tan \theta =\sqrt{\dfrac{b+a}{b-a}}$$ Let us first consider the left hand side and calculate its value $$\begin{aligned} & L.H.S=\sec \theta +\tan \theta \\\ & L.H.S=\dfrac{b}{\sqrt{{{b}^{2}}-{{a}^{2}}}}+\dfrac{a}{\sqrt{{{b}^{2}}-{{a}^{2}}}} \\\ & L.H.S=\dfrac{{{\left( \sqrt{\left( b+a \right)} \right)}^{2}}}{\sqrt{\left( b-a \right)\left( b+a \right)}} \\\ & L.H.S=\sqrt{\dfrac{b+a}{b-a}} \\\ \end{aligned}$$ Thus, the value of right hand side is equal to left hand side Hence, it is verified that $$\sec \theta +\tan \theta =\sqrt{\dfrac{b+a}{b-a}}$$ Note: Instead of calculating the values of right hand side and left hand side by substituting the respective values we can calculate either of them and then use proper trigonometric identities that both the expressions are equal. It is important to note that while calculating the values of respective functions we need to use the identities accordingly and solve them. Because neglecting any of the terms or writing it incorrectly changes the complete result.