Question

If $\sin \theta = \dfrac{{12}}{{13}}$ , find the value of $\dfrac{{{{\sin }^2}\theta - {{\cos }^2}\theta }}{{2\sin \theta \cos \theta }} \times \dfrac{1}{{{{\tan }^2}\theta }}$

Answer 1

Hint: We are provided with the value of $\sin \theta $. So we will first convert the equation in terms of $\sin \theta $by using basic trigonometric equations and then by putting the value of $\sin \theta $we’ll get the required result.



Complete step-by-step answer:
Given $\sin \theta = \dfrac{{12}}{{13}}$
$\dfrac{{{{\sin }^2}\theta - {{\cos }^2}\theta }}{{2\sin \theta \cos \theta }} \times \dfrac{1}{{{{\tan }^2}\theta }}$…………………….(1)
Converting the equation in terms of sin by replacing $\cos \theta $and $\tan \theta $ by using $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ and${\cos ^2}\theta = 1 - {\sin ^2}\theta $
then the equation (1) becomes
$ = \dfrac{{{{\sin }^2}\theta - \left( {1 - {{\sin }^2}\theta } \right)}}{{2\sin \theta \cos \theta }} \times \dfrac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }}$
$ = \dfrac{{{{\sin }^2}\theta - 1 + {{\sin }^2}\theta }}{{2\sin \theta \cos \theta }} \times \dfrac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }}$
after using simple math, we get
$ = \dfrac{{2{{\sin }^2}\theta - 1}}{{2{{\sin }^3}\theta }} \times \cos \theta $
Putting the value of $\cos \theta $=\[\sqrt {1 - {{\sin }^2}\theta } \] we get
$ = \dfrac{{2{{\sin }^2}\theta - 1}}{{2{{\sin }^3}\theta }} \times \sqrt {1 - {{\sin }^2}\theta } $
Now we have found the equation in terms of $\sin \theta $. So putting the value of $\sin \theta $ in above equation
$ = \dfrac{{2{{\left( {\dfrac{{12}}{{13}}} \right)}^2} - 1}}{{2{{\left( {\dfrac{{12}}{{13}}} \right)}^3}}} \times \sqrt {1 - {{\left( {\dfrac{{12}}{{13}}} \right)}^2}} $
On taking the LCM as \[ \left( {13} \right)^2\]
$ = \dfrac{{2{{\left( {12} \right)}^2} - {{\left( {13} \right)}^2}}}{{2 \times {{\left( {13} \right)}^2} \times {{\left( {\dfrac{{12}}{{13}}} \right)}^3}}} \times \sqrt {\dfrac{{{{\left( {13} \right)}^2} - {{\left( {12} \right)}^2}}}{{{{\left( {13} \right)}^2}}}} $
on putting the value of square of 12 and 13, we get
=$\dfrac{{288 - 169}}{{2 \times \dfrac{{{{\left( {12} \right)}^3}}}{{13}}}} \times \dfrac{{\sqrt {169 - 144} }}{{13}}$
$ = \dfrac{{119}}{{2 \times 1728}} \times \sqrt {25} $
$ = \dfrac{{119 \times 5}}{{3456}}$
$ = \dfrac{{595}}{{3456}}$
So value of $\dfrac{{{{\sin }^2}\theta - {{\cos }^2}\theta }}{{2\sin \theta \cos \theta }} \times \dfrac{1}{{{{\tan }^2}\theta }}$ is $\dfrac{{595}}{{3456}}$.

Note- Some basic trigonometric equation should be in our mind which is very useful for converting the equation
${\sin ^2}\theta + {\cos ^2}\theta = 1$
${\sec ^2}\theta - {\tan ^2}\theta = 1$
$\cos e{c^2}\theta - {\cot ^2}\theta = 1$