Question

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

Answer 1

Hint:First of all consider a right angled triangle ABC with C as an angle $\theta$. Now as $\tan \theta =\dfrac{12}{13}$, consider perpendicular and base as 12x and 13x respectively. Now use Pythagoras theorem to find the hypotenuse of a triangle. Now find $\sin \theta =\dfrac{P}{H}$ and $\cos \theta =\dfrac{B}{H}$ and substitute in the given expression to get the required answer.

Complete step-by-step answer:
Here, we are given that $\tan \theta =\dfrac{12}{13}$ and we have to find the value of $\dfrac{2\sin \theta \cos \theta }{{{\cos }^{2}}\theta -{{\sin }^{2}}\theta }$.
Let us consider the expression given in the question.
$E=\dfrac{2\sin \theta \cos \theta }{{{\cos }^{2}}\theta -{{\sin }^{2}}\theta }......(1)$
We are given that $\tan \theta =\dfrac{12}{13}......(2)$
We know that $\tan \theta =\dfrac{perpendicular}{base}.....(3)$
From equation (2) and (3) we get as follows:
$\dfrac{12}{13}=\dfrac{perpendicular}{base}$
Let us assume a $\Delta ABC$ is right angled at B and angle C is $\theta$.

Let perpendicular AB be equal to 12x and base BC be equal to 13x.
We know that Pythagoras theorem states that in a right angled triangle, the square of the hypotenuse is equal to the sum of squares of the other two sides.
So in above $\Delta ABC$, by applying Pythagoras theorem, we get as follows:
${{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}={{\left( AC \right)}^{2}}$
By substituting the value of AB = 12x and BC = 13x, we get as follows:

& {{\left( 12x \right)}^{2}}+{{\left( 13x \right)}^{2}}={{\left( AC \right)}^{2}} \\\ & 144{{x}^{2}}+168{{x}^{2}}={{\left( AC \right)}^{2}} \\\ & 313{{x}^{2}}={{\left( AC \right)}^{2}} \\\ \end{aligned}$$ So we get $$AC=\sqrt{313{{x}^{2}}}=x\sqrt{313}$$ We know that $$\sin \theta =\dfrac{perpendicular}{hypotenuse}....(4)$$ In $$\Delta ABC$$ with respect to angle $$\theta $$, Perpendicular = AB = 12x Hypotenuse = AC = 13x By substituting these values in equation (4), we get as follows: $$\sin \theta =\dfrac{12x}{\sqrt{313}x}=\dfrac{12}{\sqrt{313}}$$ We also know that $$\cos \theta =\dfrac{Base}{Hypotenuse}......(5)$$ In $$\Delta ABC$$ with respect to angle $$\theta $$, Base = BC = 13x Hypotenuse = AC = $$\sqrt{313}x$$ By substituting the values in equation (5) we get as follows: $$\cos \theta =\dfrac{13x}{\sqrt{313}x}=\dfrac{13}{\sqrt{313}}$$ Now, by substituting the values of $$\sin \theta =\dfrac{12}{\sqrt{313}}$$ and $$\cos \theta =\dfrac{13}{\sqrt{313}}$$ in equation (1), we get as follows: $$E=\dfrac{2\sin \theta \cos \theta }{{{\cos }^{2}}\theta -{{\sin }^{2}}\theta }$$ $$\begin{aligned} & E=\dfrac{2.\dfrac{12}{\sqrt{313}}.\dfrac{13}{\sqrt{313}}}{{{\left( \dfrac{13}{\sqrt{313}} \right)}^{2}}-{{\left( \dfrac{12}{\sqrt{313}} \right)}^{2}}} \\\ & E=\dfrac{\dfrac{312}{313}}{\dfrac{169}{313}-\dfrac{144}{313}} \\\ & E=\dfrac{\dfrac{312}{313}}{\dfrac{25}{313}} \\\ & E=\left( \dfrac{312}{313}\times \dfrac{313}{25} \right) \\\ \end{aligned}$$ By cancelling the like terms, we get as follows: $$E=\dfrac{312}{25}$$ So we have got the value of $$\dfrac{2\sin \theta \cos \theta }{{{\cos }^{2}}\theta -{{\sin }^{2}}\theta }$$ as $$\dfrac{312}{25}$$. Note: In this question, students must take sides as 12x and 13x etc. and not as 12 and 13 because we’re only given the ratio and not the absolute value of sides. Also students must note that in a right angled triangle, hypotenuse remains constant but the values of perpendicular and base side change according to the angle we consider.