Answer

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Hint: Use trigonometric identities to simplify the problem.

We know the half angle formula for sin and cosine. These are $\sin 2a = 2\sin a\cos a$ and $\cos 2a = {\cos ^2}a - {\sin ^2}a$ . By using these identities, $\dfrac{{2\sin x\cos x}}{{{{\cos }^2}x - {{\sin }^2}x}} = \dfrac{{\sin 2x}}{{\cos 2x}} = \tan 2x$ . Now, we have given $\tan x = \dfrac{{12}}{{13}}$ and we need to get the value of $\tan 2x$ . We can use the half angle formula again to write $\tan 2x$ in terms of $\tan x$ . We know that, $\tan 2a = \dfrac{{2\tan a}}{{1 - {{\tan }^2}a}}$ . Putting the values, we’ll get,

$

\tan 2a = \dfrac{{2\tan a}}{{1 - {{\tan }^2}a}} \\

\Rightarrow \tan 2a = \dfrac{{2\dfrac{{12}}{{13}}}}{{1 - {{(\dfrac{{12}}{{13}})}^2}}}{\text{ }}[{\text{Using}},\tan a = \dfrac{{12}}{{13}}] \\

\Rightarrow \tan 2a = \dfrac{{\dfrac{{24}}{{13}}}}{{1 - \dfrac{{144}}{{169}}}} \\

\Rightarrow \tan 2a = \dfrac{{\dfrac{{24}}{{13}}}}{{\dfrac{{169 - 144}}{{169}}}} \\

\Rightarrow \tan 2a = \dfrac{{\dfrac{{24}}{{13}}}}{{\dfrac{{25}}{{169}}}} \\

\Rightarrow \tan 2a = \dfrac{{24}}{{13}} \times \dfrac{{169}}{{25}} \\

\Rightarrow \tan 2a = \dfrac{{24}}{1} \times \dfrac{{13}}{{25}} \\

\Rightarrow \tan 2a = \dfrac{{312}}{{25}} \\

$

Hence, the required value of $\dfrac{{2\sin x\cos x}}{{{{\cos }^2}x - {{\sin }^2}x}} = \dfrac{{312}}{{25}}$ .

Note: There is more process to solve this question. We have given$\tan $ratio so we can use it to get the ratios of$\sin {\text{ and cos}}$. Then just putting the value in $\dfrac{{2\sin x\cos x}}{{{{\cos }^2}x - {{\sin }^2}x}}$ will give us the answer.

We know the half angle formula for sin and cosine. These are $\sin 2a = 2\sin a\cos a$ and $\cos 2a = {\cos ^2}a - {\sin ^2}a$ . By using these identities, $\dfrac{{2\sin x\cos x}}{{{{\cos }^2}x - {{\sin }^2}x}} = \dfrac{{\sin 2x}}{{\cos 2x}} = \tan 2x$ . Now, we have given $\tan x = \dfrac{{12}}{{13}}$ and we need to get the value of $\tan 2x$ . We can use the half angle formula again to write $\tan 2x$ in terms of $\tan x$ . We know that, $\tan 2a = \dfrac{{2\tan a}}{{1 - {{\tan }^2}a}}$ . Putting the values, we’ll get,

$

\tan 2a = \dfrac{{2\tan a}}{{1 - {{\tan }^2}a}} \\

\Rightarrow \tan 2a = \dfrac{{2\dfrac{{12}}{{13}}}}{{1 - {{(\dfrac{{12}}{{13}})}^2}}}{\text{ }}[{\text{Using}},\tan a = \dfrac{{12}}{{13}}] \\

\Rightarrow \tan 2a = \dfrac{{\dfrac{{24}}{{13}}}}{{1 - \dfrac{{144}}{{169}}}} \\

\Rightarrow \tan 2a = \dfrac{{\dfrac{{24}}{{13}}}}{{\dfrac{{169 - 144}}{{169}}}} \\

\Rightarrow \tan 2a = \dfrac{{\dfrac{{24}}{{13}}}}{{\dfrac{{25}}{{169}}}} \\

\Rightarrow \tan 2a = \dfrac{{24}}{{13}} \times \dfrac{{169}}{{25}} \\

\Rightarrow \tan 2a = \dfrac{{24}}{1} \times \dfrac{{13}}{{25}} \\

\Rightarrow \tan 2a = \dfrac{{312}}{{25}} \\

$

Hence, the required value of $\dfrac{{2\sin x\cos x}}{{{{\cos }^2}x - {{\sin }^2}x}} = \dfrac{{312}}{{25}}$ .

Note: There is more process to solve this question. We have given$\tan $ratio so we can use it to get the ratios of$\sin {\text{ and cos}}$. Then just putting the value in $\dfrac{{2\sin x\cos x}}{{{{\cos }^2}x - {{\sin }^2}x}}$ will give us the answer.

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