If tan θ=17√, show that cosec2 θ−sec2 θcosec2 θ+sec2 θ=34

Question:

If $\tan \theta=\frac{1}{\sqrt{7}}$, show that $\frac{\operatorname{cosec}^{2} \theta-\sec ^{2} \theta}{\operatorname{cosec}^{2} \theta+\sec ^{2} \theta}=\frac{3}{4}$

Solution:

Given: $\tan \theta=\frac{1}{\sqrt{7}}$....(1)

To show that $\frac{\operatorname{cosec}^{2} \theta-\sec ^{2} \theta}{\operatorname{cosec}^{2} \theta+\sec ^{2} \theta}=\frac{3}{4}$

Now, we know that

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

Therefore,

Comparing Equation (1) and (2)

We get,

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

Base side adjacent to $\angle \theta=\sqrt{7}$

Therefore, Triangle representing angle $\theta$ is as shown below

Hypotenuse AC is unknown and it can be found by using Pythagoras theorem

Therefore by applying Pythagoras theorem

We get,

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

Therefore by substituting the values of known sides

We get,

$A C^{2}=(1)^{2}+(\sqrt{7})^{2}$

Therefore,

$A C^{2}=1+7$

$A C^{2}=8$

$A C=\sqrt{8}$

$A C=\sqrt{2 \times 2 \times 2}$

Therefore,

$A C=2 \sqrt{2} \ldots \ldots .(3)$

Now, we know that

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

Now from figure (a)

We get,

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

Therefore from figure (a) and equation (3) ,

$\sin \theta=\frac{1}{2 \sqrt{2}}$...(4)

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

Therefore,

$\operatorname{cosec} \theta=2 \sqrt{2} \ldots \ldots(5)$

Now, we know that

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

Now from figure (a)

We get,

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

Therefore from figure (a) and equation (3) ,

$\cos \theta=\frac{\sqrt{7}}{2 \sqrt{2}}$....(6)

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

Therefore, from equation (6)

We get,

$\sec \theta=\frac{1}{\frac{\sqrt{7}}{2 \sqrt{2}}}$

Therefore,

$\sec \theta=\frac{2 \sqrt{2}}{\sqrt{7}} \ldots \ldots$....(7)

Now, L.H.S of the equation to be proved is as follows

L.H.S $=\frac{\operatorname{cosec}^{2} \theta-\sec ^{2} \theta}{\operatorname{cosec}^{2} \theta+\sec ^{2} \theta}$

Substituting the value of $\operatorname{cosec} \theta$ and $\sec \theta$ from equation $(6)$ and $(7)$

We get,

L.H.S $=\frac{\left[(2 \sqrt{2})^{2}\right]-\left(\frac{2 \sqrt{2}}{\sqrt{7}}\right)^{2}}{\left[(2 \sqrt{2})^{2}\right]+\left(\frac{2 \sqrt{2}}{\sqrt{7}}\right)^{2}}$

L.H.S $=\frac{(4 \times 2)-\frac{(4 \times 2)}{7}}{(4 \times 2)+\frac{(4 \times 2)}{7}}$

$L . H . S=\frac{(8)-\frac{(8)}{7}}{(8)+\frac{(8)}{7}}$

Now by taking L.C.M. in numerator as well as denominator

We get,

L.H.S $=\frac{\frac{(7 \times 8)-8}{7}}{\frac{(7 \times 8)+8}{7}}$

Therefore,

$L . H . S=\frac{\frac{56-8}{7}}{\frac{56+8}{7}}$

L.H.S $=\frac{\frac{48}{7}}{\frac{64}{7}}$

Therefore,

L.H.S $=\frac{48}{7} \times \frac{7}{64}$

$L . H . S=\frac{48}{64}$

L.H.S $=\frac{3}{4}$

$L . H . S=\frac{3}{4}=R . H . S$

Therefore,

$\frac{\operatorname{cosec}^{2} \theta-\sec ^{2} \theta}{\operatorname{cosec}^{2} \theta+\sec ^{2} \theta}=\frac{3}{4}$

Hence proved that

$\frac{\operatorname{cosec}^{2} \theta-\sec ^{2} \theta}{\operatorname{cosec}^{2} \theta+\sec ^{2} \theta}=\frac{3}{4}$