If $ \sin \theta=\frac{a}{b} $, find $ \sec \theta+\tan \theta $ in terms of $ a $ and $ b $

Given:

\( \sin \theta=\frac{a}{b} \).

To do:

We have to find the value of \( \sec \theta+\tan \theta \).

Solution:  

Let, in a triangle $ABC$ right-angled at $B$, $\ sin\ \theta = sin\ A = \frac{a}{b}$.

We know that,

In a right-angled triangle $ABC$ with right angle at $B$,

By Pythagoras theorem,

$AC^2=AB^2+BC^2$

By trigonometric ratios definitions,

$sec\ \theta=\frac{Hypotenuse}{Adjacent}=\frac{AC}{AB}$

$tan\ \theta=\frac{Opposite}{Adjacent}=\frac{BC}{AB}$

Here,

$AC^2=AB^2+BC^2$

$\Rightarrow (b)^2=AB^2+(a)^2$

$\Rightarrow AB^2=b^2-a^2$

$\Rightarrow AB=\sqrt{b^2-a^2}$

Therefore,

$sec\ \theta=\frac{AC}{AB}=\frac{b}{\sqrt{b^2-a^2}}$

$tan\ \theta=\frac{BC}{AB}=\frac{a}{\sqrt{b^2-a^2}}$

 $\sec \theta+\tan \theta= \frac{b}{\sqrt{b^2-a^2}}+\frac{a}{\sqrt{b^2-a^2}}$

$=\frac{a+b}{\sqrt{(b+a)(b-a)}}$

$=\frac{a+b}{\sqrt{(b+a)(b-a)}}\times\frac{\sqrt{b+a}}{\sqrt{(b+a)}}$

$=\frac{(a+b)\times\sqrt{b+a}}{\sqrt{(b+a)(b-a)}\times\sqrt{(b+a)}}$

$=\frac{(a+b)\times\sqrt{b+a}}{(a+b)\times\sqrt{(b-a)}}$

$=\sqrt{\frac{b+a}{b-a}}$

The value of $\sec \theta+\tan \theta$ is $\sqrt{\frac{b+a}{b-a}}$. 

Tutorialspoint

Updated on: 10-Oct-2022

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