Question
If $\sin\theta=\frac{\text{a}}{\text{b}},$ show that $(\sec\theta+\tan\theta)=\sqrt{\frac{\text{b}+\text{a}}{\text{b}-\text{a}}}.$
✓
Answer
$\sin\theta=\frac{\text{a}}{\text{b}}$
Consider $\triangle\text{ABC},$ where $\angle\text{B}=90^\circ$ and $\angle\text{A}=\theta$
Then, $\sin\theta=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{\text{a}}{\text{b}}$
Let BC = a and AC = b
Then, by pythagoras theorem,
$\text{AB}^2 = \text{AC}^2 - \text{BC}^2 = \text{b}^2 - \text{a}^2$
$\Rightarrow\text{AB}=\sqrt{\text{b}^2-\text{a}^2}$
Now,
$\sec\theta=\frac{\text{Hypotenuse}}{\text{Base}}=\frac{\text{AC}}{\text{AB}}=\frac{\text{b}}{\sqrt{\text{b}^2-\text{a}^2}}$
$\tan\theta=\frac{\text{Perpendicular}}{\text{Base}}=\frac{\text{BC}}{\text{AB}}=\frac{\text{a}}{\sqrt{\text{b}^2-\text{a}^2}}$
$\therefore\text{L.H.S.}=(\sec\theta+\tan\theta)$
$=\frac{\text{b}}{\sqrt{\text{b}^2-\text{a}^2}}+\frac{\text{a}}{\sqrt{\text{b}^2-\text{a}^2}}$
$=\frac{\text{b}+\text{a}}{\sqrt{\text{b}^2-\text{a}^2}}$
$\frac{\text{b}+\text{a}}{\sqrt{(\text{b}+\text{a})(\text{b}-\text{a})}}$
$=\frac{{\sqrt{\text{b}+\text{a}}}\times\sqrt{\text{b}+\text{a}}}{\sqrt{(\text{b}+\text{a})}\times\sqrt{\text{b}-\text{a}}}$
$=\sqrt{\frac{\text{b}+\text{a}}{\text{b}-\text{a}}}$
$=\text{R.H.S.}$
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