Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!

If tan θ=1213, find the value of 2 sin θ cos θcos2 θ−sin2 θ

Question:

If $\tan \theta=\frac{12}{13}$, find the value of $\frac{2 \sin \theta \cos \theta}{\cos ^{2} \theta-\sin ^{2} \theta}$

Solution:

Given: $\tan \theta=\frac{12}{13}$

To find the value of $\frac{2 \sin \theta \cos \theta}{\cos ^{2} \theta-\sin ^{2} \theta}$

Now, we know the following trigonometric identity

$\operatorname{cosec}^{2} \theta=1+\tan ^{2} \theta$

Therefore, by substituting the value of $\tan \theta$ from equation (1),

We get,

$\operatorname{cosec}^{2} \theta=1+\left(\frac{12}{13}\right)^{2}$

$=1+\frac{(12)^{2}}{(13)^{2}}$

$=1+\frac{144}{169}$

By taking L.C.M. on the R.H.S,

We get,

$\operatorname{cosec}^{2} \theta=\frac{169+144}{169}$

$=\frac{313}{169}$

Therefore

$\operatorname{cosec} \theta=\frac{\sqrt{313}}{13}$.....(2)

Now, we know that

$\operatorname{cosec} \theta=\frac{1}{\sin \theta}$

Therefore,

$\sin \theta=\frac{1}{\operatorname{cosec} \theta}$

$\sin \theta=\frac{1}{\frac{\sqrt{313}}{13}}$

$=\frac{13}{\sqrt{313}}$

Therefore

$\sin \theta=\frac{13}{\sqrt{313}}$.....(3)

Now, we know the following trigonometric identity

$\cos ^{2} \theta+\sin ^{2} \theta=1$

Now by substituting the value of $\sin \theta$ from equation (3)

We get,

$\cos ^{2} \theta=1-\left(\frac{13}{\sqrt{313}}\right)^{2}$

$=1-\frac{(13)^{2}}{(\sqrt{313})^{2}}$

$=1-\frac{169}{313}$

$=1-\frac{169}{313}$

Therefore, by taking L.C.M on R.H.S

We get,

$\cos ^{2} \theta=\frac{313-169}{313}$

$=\frac{144}{313}$

Now, by taking square root on both sides

We get,

$\cos \theta=\sqrt{\frac{144}{313}}$

$=\frac{12}{\sqrt{313}}$

Therefore,

$\cos \theta=\frac{12}{\sqrt{313}}$....(4)

Substituting the value of $\sin \theta$ and $\cos \theta$ from equation (3) and (4) respectively in the expression below

$\frac{2 \sin \theta \cos \theta}{\cos ^{2} \theta-\sin ^{2} \theta}$

Therefore,

$\frac{2 \sin \theta \cos \theta}{\cos ^{2} \theta-\sin ^{2} \theta}=\frac{2 \times \frac{13}{\sqrt{313}} \times \frac{12}{\sqrt{313}}}{\left(\frac{13}{\sqrt{313}}\right)^{2}-\left(\frac{12}{\sqrt{313}}\right)^{2}}$

$=\frac{\frac{2 \times 13 \times 12}{313}}{\frac{(13)^{2}}{313}-\frac{(12)^{2}}{313}}$

$=\frac{\frac{312}{313}}{\frac{169}{313}-\frac{144}{313}}$

$=\frac{\frac{312}{313}}{\frac{169-144}{313}}$

$=\frac{\frac{312}{313}}{\frac{25}{313}}$

$=\frac{312}{25}$

Therefore, $\frac{2 \sin \theta \cos \theta}{\cos ^{2} \theta-\sin ^{2} \theta}=\frac{312}{25}$