Question

If the value the trigonometric ratio $\sin \theta =\dfrac{3}{4}$, prove that
$\sqrt{\dfrac{{{\operatorname{cosec}}^{2}}\theta -{{\cot }^{2}}\theta }{{{\sec }^{2}}\theta -1}}=\dfrac{\sqrt{7}}{3}$

Answer 1

First of all, consider a triangle ABC, with C as the angle $\theta$. Now as $\sin \theta =\dfrac{3}{4}$. So, consider perpendicular and hypotenuse at 3x and 4x respectively. Now, find the remaining side by using the Pythagoras theorem. Now, find $\operatorname{cosec}\theta ,\cot \theta \text{ and }\sec \theta$ from this triangle and then substitute in the LHS to prove the desired result.

Complete step-by-step answer:
Here, we are given that $\sin \theta =\dfrac{3}{4}$. We have to prove that $\sqrt{\dfrac{{{\operatorname{cosec}}^{2}}\theta -{{\cot }^{2}}\theta }{{{\sec }^{2}}\theta -1}}=\dfrac{\sqrt{7}}{3}$.
We are given that,
$\sin \theta =\dfrac{3}{4}....\left( i \right)$
We know that,
$\sin \theta =\dfrac{Perpendicular}{Hypotenuse}....\left( ii \right)$
From equation (i) and (ii), we get,
$\dfrac{3}{4}=\dfrac{Perpendicular}{Hypotenuse}$
Let us consider a triangle ABC, right-angled at B and angle C is $\theta$.

Let perpendicular AB be equal to 3x and hypotenuse be equal to 4x.
We know that Pythagoras theorem states that in a right-angled triangle, the square of the hypotenuse side is equal to the sum of the squares of the other two sides. So, in the above triangle ABC, by applying Pythagoras theorem, we get,
${{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}={{\left( AC \right)}^{2}}$
Now by substituting the value of AB = 3x and AC = 4x, we get,
${{\left( 3x \right)}^{2}}+{{\left( BC \right)}^{2}}={{\left( 4x \right)}^{2}}$
$9{{x}^{2}}+{{\left( BC \right)}^{2}}=16{{x}^{2}}$
$B{{C}^{2}}=16{{x}^{2}}-9{{x}^{2}}$
$B{{C}^{2}}=7{{x}^{2}}$
$BC=\sqrt{7}x$
Now, we know that,
$\operatorname{cosec}\theta =\dfrac{Hypotenuse}{Perpendicular}$
$\cot \theta =\dfrac{Base}{Perpendicular}$
$\sec \theta =\dfrac{Hypotenuse}{Base}$
We can see that with respect to angle $\theta$,
Perpendicular = AB = 3x
Base = BC = $\sqrt{7}x$
Hypotenuse = AC = 4x
So, we get,
$\operatorname{cosec}\theta =\dfrac{AC}{AB}=\dfrac{4x}{3x}=\dfrac{4}{3}$
$\cot \theta =\dfrac{BC}{AB}=\dfrac{\sqrt{7}x}{3x}=\dfrac{\sqrt{7}}{3}$
$\sec \theta =\dfrac{AC}{BC}=\dfrac{4x}{\sqrt{7}x}=\dfrac{4}{\sqrt{7}}$
Now, let us consider the LHS of the equation given in the question,
$LHS=\sqrt{\dfrac{{{\operatorname{cosec}}^{2}}\theta -{{\cot }^{2}}\theta }{{{\sec }^{2}}\theta -1}}$
By substituting the value of $\operatorname{cosec}\theta ,\cot \theta \text{ and }\sec \theta$, we get,
$LHS=\sqrt{\dfrac{{{\left( \dfrac{4}{3} \right)}^{2}}-{{\left( \dfrac{\sqrt{7}}{3} \right)}^{2}}}{{{\left( \dfrac{4}{\sqrt{7}} \right)}^{2}}-1}}$
$LHS=\sqrt{\dfrac{\dfrac{16}{9}-\dfrac{7}{9}}{\dfrac{16}{7}-1}}$
$LHS=\sqrt{\dfrac{\dfrac{9}{9}}{\dfrac{9}{7}}}$
$LHS=\sqrt{\dfrac{1}{\dfrac{9}{7}}}$
$LHS=\sqrt{\dfrac{7}{9}}$
$LHS=\dfrac{\sqrt{7}}{3}=RHS$
Hence proved
So, we have proved that
$\sqrt{\dfrac{{{\operatorname{cosec}}^{2}}\theta -{{\cot }^{2}}\theta }{{{\sec }^{2}}\theta -1}}=\dfrac{\sqrt{7}}{3}$

Note: Students can also solve this question in the following way,
$LHS=\sqrt{\dfrac{{{\operatorname{cosec}}^{2}}\theta -{{\cot }^{2}}\theta }{{{\sec }^{2}}\theta -1}}$
We know that $1+{{\cot }^{2}}\theta ={{\operatorname{cosec}}^{2}}\theta$ or ${{\operatorname{cosec}}^{2}}\theta -{{\cot }^{2}}\theta =1$. Also, we know that ${{\sec }^{2}}\theta -1={{\tan }^{2}}\theta$. By using these, we get,
$LHS=\sqrt{\dfrac{1}{{{\tan }^{2}}\theta }}=\dfrac{1}{\tan \theta }$
We know that,
$\tan \theta =\dfrac{P}{B}=\dfrac{3}{\sqrt{7}}$
So, we get,
$LHS=\dfrac{1}{\dfrac{3}{\sqrt{7}}}=\dfrac{\sqrt{7}}{3}=RHS$
Hence proved