If sin θ=34, prove that cosec2 θ−cot2 θsec2 θ−1−−−−−−−−−−√=7√3.

Solution:

Given:

$\sin \theta=\frac{3}{4} \ldots \ldots$(1)

To prove:

$\sqrt{\frac{\operatorname{cosec}^{2} \theta-\cot ^{2} \theta}{\sec ^{2} \theta-1}}=\frac{\sqrt{7}}{3}$.....(2)

By definition,

$\sin \mathrm{A}=\frac{\text { Perpendicular side opposite to } \angle A}{\text { Hypotenuse }}$.....(3)

By Comparing (1) and (3)

We get,

Perpendicular side = 3 and

Hypotenuse = 4

Side BC is unknown.

So we find $\mathrm{BC}$ by applying Pythagoras theorem to right angled $\triangle A B C$,

Hence,

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

Now we substitute the value of perpendicular side (AB) and hypotenuse (AC) and get the base side (BC)

Therefore,

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

$B C^{2}=4^{2}-3^{2}$

$B C^{2}=16-9$

$B C^{2}=7$

$B C=\sqrt{7}$

$B C^{2}=4^{2}-3^{2}$

$B C^{2}=16-9$

$B C^{2}=7$

$B C=\sqrt{7}$

Hence, Base side BC $=\sqrt{7}$.....(3)

Now, $\cos A=\frac{\text { Base }}{\text { Hypotenuse }}$

Therefore from fig. a and equation (3)

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

$=\frac{\sqrt{7}}{4}$

Therefore,

$\cos A=\frac{\sqrt{7}}{4}$....(4)

Now, $\operatorname{cosec} A=\frac{1}{\sin A}$

Therefore from fig. a and equation (1) ,

$\operatorname{cosec} A=\frac{\text { Hypotenuse }}{\text { Perpendicular }}$

$\operatorname{cosec} A=\frac{4}{3}$.....(5)

Now, $\sec A=\frac{1}{\cos \mathrm{A}}$

Therefore from fig. a and equation (4) ,

$\sec A=\frac{4}{\sqrt{7}}$.....(6)

Now, $\cot A=\frac{\cos \mathrm{A}}{\sin A}$

Therefore by substituting the values from equation (1) and (4) ,

We get,

$\cot A=\frac{\frac{\sqrt{7}}{4}}{\frac{3}{4}}$'

$=\frac{\sqrt{7}}{4} \times \frac{4}{3}$

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

Therefore,

$\cot A=\frac{\sqrt{7}}{3}$....(7)

Now by substituting the value of $\operatorname{cosec} A, \sec A$ and $\cot A$ from equation $(5),(6)$ and (7) respectively in the L. H.S of expression (2),

 

We get,

$\sqrt{\frac{\operatorname{cosec}^{2} \theta-\cot ^{2} \theta}{\sec ^{2} \theta-1}}=\sqrt{\frac{\left(\frac{4}{3}\right)^{2}-\left(\frac{\sqrt{7}}{3}\right)^{2}}{\left(\frac{4}{\sqrt{7}}\right)^{2}-1}}$

$=\sqrt{\frac{\frac{(4)^{2}}{(3)^{2}}-\frac{(\sqrt{7})^{2}}{(3)^{2}}}{\frac{(4)^{2}}{(\sqrt{7})^{2}}-1}}$

$=\sqrt{\frac{\frac{16}{9}-\frac{7}{9}}{\frac{16}{7}-1}}$

$=\sqrt{\frac{\frac{16-7}{9}}{\frac{16-7}{7}}}$

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

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

Therefore,

$\sqrt{\frac{\operatorname{cosec}^{2} \theta-\cot ^{2} \theta}{\sec ^{2} \theta-1}}=\sqrt{\frac{7}{9}}$

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

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

Hence it is proved that

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