If $\sin \theta=\frac{12}{13}$, find the value of $\frac{\sin ^{2} \theta-\cos ^{2} \theta}{2 \sin \theta \cos \theta} \times \frac{1}{\tan ^{2} \theta}$
Given: $\sin \theta=\frac{12}{13}$
To Find: The value of expression $\frac{\sin ^{2} \theta-\cos ^{2} \theta}{2 \sin \theta \cos \theta} \times \frac{1}{\tan ^{2} \theta}$
Now, we know that
$\sin \theta=\frac{\text { Perpendicular side opposite to } \angle \theta}{\text { Hypotenuse }}$....(2)
Now when we compare equation (1) and (2)
We get,
Perpendicular side opposite to $\angle \theta=12$
And
Hypotenuse = 13
Therefore, Triangle representing angle $\theta$ is as shown below
Base side BC 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,
$13^{2}=12^{2}+B C^{2}$
Therefore,
$B C^{2}=13^{2}-12^{2}$
$B C^{2}=169-144$
$B C^{2}=25$
$B C=\sqrt{25}$
Therefore,
$B C=5 \ldots \ldots(3)$
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) ,
$\cos \theta=\frac{5}{13}$....(4)
Now we know that,
$\tan \theta=\frac{\sin \theta}{\cos \theta}$
Therefore, substituting the value of $\sin \theta$ and $\cos \theta$ from equation (1) and (4)
We get,
$\tan \theta=\frac{\frac{12}{13}}{\frac{5}{13}}$
$\tan \theta=\frac{12}{13} \times \frac{15}{5}$
Therefore 13 gets cancelled and we get
$\tan \theta=\frac{12}{5}$...(5)
Now we substitute the value of $\sin \theta, \cos \theta$ and $\tan \theta$ from equation (1), (4) and (5) respectively in the expression below
$\frac{\sin ^{2} \theta-\cos ^{2} \theta}{2 \sin \theta \cos \theta} \times \frac{1}{\tan ^{2} \theta}$
Therefore,
We get,
$\frac{\sin ^{2} \theta-\cos ^{2} \theta}{2 \sin \theta \cos \theta} \times \frac{1}{\tan ^{2} \theta}=\frac{\left(\frac{12}{13}\right)^{2}-\left(\frac{5}{13}\right)^{2}}{2 \times\left(\frac{12}{13}\right) \times\left(\frac{5}{13}\right)} \times \frac{1}{\left(\frac{12}{5}\right)^{2}}$
Therefore by further simplifying we get,
$\frac{\sin ^{2} \theta-\cos ^{2} \theta}{2 \sin \theta \cos \theta} \times \frac{1}{\tan ^{2} \theta}=\frac{\frac{(12)^{2}}{(13)^{2}}-\frac{(5)^{2}}{(13)^{2}}}{2 \times\left(\frac{12}{13}\right) \times\left(\frac{5}{13}\right)} \times \frac{1}{\frac{(12)^{2}}{(5)^{2}}}$
$=\frac{\frac{144}{169}-\frac{25}{169}}{\frac{2 \times 12 \times 5}{13 \times 13}} \times \frac{25}{144}$
$=\frac{\frac{144-25}{169}}{\frac{120}{169}} \times \frac{25}{144}$
$=\frac{119}{169} \times \frac{169}{120} \times \frac{25}{144}$
Now 169 gets cancelled and $\frac{25}{120}$ gets reduced to $\frac{5}{24}$
Therefore
$\frac{\sin ^{2} \theta-\cos ^{2} \theta}{2 \sin \theta \cos \theta} \times \frac{1}{\tan ^{2} \theta}=\frac{119}{1} \times \frac{1}{24} \times \frac{5}{144}$
$=\frac{119 \times 5}{24 \times 144}$
$=\frac{595}{3456}$
Therefore the value of $\frac{\sin ^{2} \theta-\cos ^{2} \theta}{2 \sin \theta \cos \theta} \times \frac{1}{\tan ^{2} \theta}$ is $\frac{595}{3456}$
That is $\frac{\sin ^{2} \theta-\cos ^{2} \theta}{2 \sin \theta \cos \theta} \times \frac{1}{\tan ^{2} \theta}=\frac{595}{3456}$