Question

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

Answer 1

$sin\theta =\frac{a}{b}$

We know that,

$\sin \theta =\frac{OppositeSide}{Hypotenuse}$

Consider the figure above,

From Pythagoras theorem,

$Hypotenus{e}^2=(Adjacentside{)}^2+(Oppositeside{)}^2$

$b^2=(Adjacentside{)}^2+a^2$

$(Adjacentside{)}^2=b^2-a^2$

$Adjacentside=\sqrt{b^2-a^2}$

$\sec \theta =\frac{Hypotenuse}{Adjacentside}=\frac{b}{\sqrt{b^2-a^2}}$

$\tan \theta =\frac{OppositeSide}{Adjacentside}=\frac{a}{\sqrt{b^2-a^2}}$

$\therefore$ $\sec \theta +\tan \theta$

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

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

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

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

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

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