Question

If $\sin \theta =\frac{3}{4}$, prove that $\sqrt{\frac{co{\sec }^2\theta -{\cot }^2\theta }{{\sec }^2\theta -1}}=\frac{\sqrt{7}}{3}$

Answer 1

Given, $\sin \theta =\frac{3}{4}$

We know that,

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

$(Hypotenuse{)}^2=(oppositeSide{)}^2+(adjacentSide{)}^2$

$4^2=3^2+(adjacentSide{)}^2$

$(adjacentSide{)}^2=16-9=7$

$\left(adjacentSide\right)=\sqrt{7}$

$\tan \theta =\frac{oppositeSide}{adjacentside}=\frac{3}{\sqrt{7}}$

To prove that $\sqrt{\frac{co{\sec }^2\theta -{\cot }^2\theta }{{\sec }^2\theta -1}}=\frac{\sqrt{7}}{3}$

Squaring on both sides,

${\left(\sqrt{\frac{co{\sec }^2\theta -{\cot }^2\theta }{{\sec }^2\theta -1}}\right)}^2={\left(\frac{\sqrt{7}}{3}\right)}^2$

$\frac{co{\sec }^2\theta -{\cot }^2\theta }{{\sec }^2\theta -1}=\frac{7}{9}$

We have,

$1+{\cot }^2\theta ={\csc }^2\theta$

$1+{\tan }^2\theta ={\sec }^2\theta$

$\frac{1}{{\tan }^2\theta }=\frac{7}{9}$

$\frac{1}{{\left(\frac{3}{\sqrt{7}}\right)}^2}=\frac{7}{9}$

$\frac{7}{9}=\frac{7}{9}$

Taking square root on both sides,

$\frac{\sqrt{7}}{3}=\frac{\sqrt{7}}{3}$

$\therefore \sqrt{\frac{co{\sec }^2\theta -{\cot }^2\theta }{{\sec }^2\theta -1}}=\frac{\sqrt{7}}{3}$

Hence Proved