Question

If $sin\theta =\frac{a}{b},$ then find the value of $(sec\theta +tan\theta )$.

• A. $\frac{b+a}{\sqrt{b^2-a^2}}$
• B. $\frac{b-a}{\sqrt{b^2-a^2}}$
• C. $\frac{b-a}{\sqrt{b^2+a^2}}$
• D. $\frac{b+a}{\sqrt{b^2+a^2}}$

Answer 1

The correct option is A $\frac{b+a}{\sqrt{b^2-a^2}}$

To find: $sec\theta +tan\theta$

We know that,
$sec\theta =\frac{1}{cos\theta }$ and $tan\theta =\frac{sin\theta }{cos\theta }$
$=\frac{1}{cos\theta }+\frac{sin\theta }{cos\theta }=\frac{1+sin\theta }{cos\theta }$

Given that $sin\theta =\frac{a}{b}$
But, $sin\theta =\frac{Oppositeside}{Hypotenuse}=\frac{a}{b}$

In a right triangle ABC,
By Pythagoras Thorem,
$(AC{)}^2=(AB{)}^2+(BC{)}^2{b}^2=a^2+(BC{)}^2(BC{)}^2=b^2-a^{2}BC=\sqrt{b^2-a^2}$

Then, $cos\theta =\frac{Adjacentside}{Hypotenuse}=\frac{\sqrt{b^2-a^2}}{b}$

So, $sec\theta +cos\theta =\frac{1+sin\theta }{cos\theta }$
By substituting the values of $sin\theta$ and $cos\theta$, we get,

$=\frac{1+\frac{a}{b}}{\frac{\sqrt{b^2-a^2}}{b}}=\frac{\frac{b+a}{b}}{\frac{\sqrt{b^2-a^2}}{b}}=\frac{b+a}{\sqrt{b^2-a^2}}$