# If cot θ=34, prove that sec θ−cosec θsec θ+cosec θ−−−−−−−−−√=17√.

Question:

If $\cot \theta=\frac{3}{4}$, prove that $\sqrt{\frac{\sec \theta-\operatorname{cosec} \theta}{\sec \theta+\operatorname{cosec} \theta}}=\frac{1}{\sqrt{7}}$

Solution:

Given:

$\cot \theta=\frac{3}{4}$.....(1)

To prove:

$\sqrt{\frac{\sec \theta-\operatorname{cosec} \theta}{\sec \theta+\operatorname{cosec} \theta}}=\frac{1}{\sqrt{7}}$

Now we know $\tan \theta$ is defined as follows

$\cot \theta=\frac{\text { Base side adjacent to } \angle \theta}{\text { Perpendicular side opposite to } \angle \theta}$.....(2)

Now by comparing equation (1) and (2)

We get,

Base side adjacent to $\angle \theta=3$

Perpendicular side opposite to $\angle \theta=4$

Therefore triangle representing angle $\theta$ is as shown below

Side AC is unknown and can be found using Pythagoras theorem

Therefore,

$A C^{2}=A B^{2}+B C^{2}$

Now by substituting the value of known sides from figure

We get,

$A C^{2}=4^{2}+3^{2}$

$=16+9$

$=25$

Now by taking square root on both sides

We get,

$A C=\sqrt{25}$

$=5$

Therefore Hypotenuse side AC = 5 …… (3)

Now we know, $\sin \theta$ is defined as follows

$\sin \theta=\frac{\text { Perpendicular side opposite to } \angle \theta}{\text { Hypotenuse }}$

Therefore from figure (a) and equation (3)

We get,

$\sin \theta=\frac{A B}{A C}$

$=\frac{4}{5}$

$\sin \theta=\frac{4}{5} \ldots \ldots$(4)

Now we know $\operatorname{cosec} \theta=\frac{1}{\sin \theta}$

Therefore by substituting the value of $\sin \theta$ from equation (4)

We get,

$\operatorname{cosec} \theta=\frac{1}{\frac{4}{5}}$

$=\frac{5}{4}$

Therefore,

$\operatorname{cosec} \theta=\frac{5}{4}$....(5)

Now we know, $\cos \theta$ is defined as follows

$\cos \theta=\frac{\text { Base side adjacent to } \angle \theta}{\text { Hypotenuse }}$

Therefore from figure (a) and equation (3)

We get,

$\cos \theta=\frac{B C}{A C}$

$=\frac{3}{5}$

$\cos \theta=\frac{3}{5} \ldots \ldots$(6)

Now we know $\sec \theta=\frac{1}{\cos \theta}$

Therefore by substituting the value of $\cos \theta$ from equation (6)

We get,

$\sec \theta=\frac{1}{\frac{3}{5}}$

$=\frac{5}{3}$

Therefore,

$\sec \theta=\frac{5}{3} \ldots \ldots(7)$

Now, in expression $\sqrt{\frac{\sec \theta-\operatorname{cosec} \theta}{\sec \theta+\operatorname{cosec} \theta}}$, by substituting the value of $\operatorname{cosec} \theta$ and $\sec \theta$ from equation (6) and (7) respectively, we get,

$\sqrt{\frac{\sec \theta-\operatorname{cosec} \theta}{\sec \theta+\operatorname{cosec} \theta}}=\sqrt{\frac{\frac{5}{3}-\frac{5}{4}}{\frac{5}{3}+\frac{5}{4}}}$

L.C.M of 3 and 4 is 12

Now by taking L.C.M in above expression

We get,

$\sqrt{\frac{\sec \theta-\operatorname{cosec} \theta}{\sec \theta+\operatorname{cosec} \theta}}=$\sqrt{\frac{\frac{5 \times 4}{3 \times 4}-\frac{5 \times 3}{4 \times 3}}{\frac{5 \times 4}{3 \times 4}+\frac{5 \times 3}{4 \times 3}}}=\sqrt{\frac{\frac{20}{12}-\frac{15}{12}}{\frac{20}{12}+\frac{15}{12}}}=\sqrt{\frac{\frac{20-15}{12}}{\frac{20+15}{12}}}=\sqrt{\frac{\frac{5}{12}}{\frac{35}{12}}}=\sqrt{\frac{5}{12} \times \frac{12}{35}}$Now 12 gets cancelled and we get,$\sqrt{\frac{\sec \theta-\operatorname{cosec} \theta}{\sec \theta+\operatorname{cosec} \theta}}=\sqrt{\frac{5}{35}}$Now$35=5 \times 7$Therefore,$\sqrt{\frac{\sec \theta-\operatorname{cosec} \theta}{\sec \theta+\operatorname{cosec} \theta}}=\sqrt{\frac{5}{5 \times 7}}$Now 5 gets cancelled and we get,$\sqrt{\frac{\sec \theta-\operatorname{cosec} \theta}{\sec \theta+\operatorname{cosec} \theta}}=\sqrt{\frac{1}{7}}=\frac{1}{\sqrt{7}}$Therefore, it is proved that$\sqrt{\frac{\sec \theta-\operatorname{cosec} \theta}{\sec \theta+\operatorname{cosec} \theta}}=\frac{1}{\sqrt{7}}\$