Question

If $\sin \theta = \dfrac{{12}}{{13}}$ , how do you 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 asked to calculate the value of the given expression. The value of sine function is given, try to express sine of an angle in terms of sides of a right angled triangle. Using Pythagoras theorem, find the values of all sides of the right angled triangle and use these values to find the value of the cosine and tangent function. Then use these values to find the value of the given expression.

Complete step-by-step answer :
Given, $\sin \theta = \dfrac{{12}}{{13}}$
We are asked to find the value of $\dfrac{{{{\sin }^2}\theta - {{\cos }^2}\theta }}{{2\sin \theta \cos \theta }} \times \dfrac{1}{{{{\tan }^2}\theta }}$ .
To calculate this we need to find the values of $\cos \theta$ and $\tan \theta$ .
Sine of an angle can be written in terms of perpendicular and hypotenuse of a right angled triangle. That is we can write,
$\sin \theta = \dfrac{{{\text{Perpendicular}}}}{{{\text{Hypotenuse}}}}$ (i)
Similarly cosine of an angle can also be written in terms of base and hypotenuse of a right angled triangle. That is,
$\cos \theta = \dfrac{{{\text{Base}}}}{{{\text{Hypotenuse}}}}$ (ii)
Suppose $\theta$ is an angle of a right angled triangle ABC such that $\angle B = \theta$ .

We are given $\sin \theta = \dfrac{{12}}{{13}}$ , comparing this value with equation (i) we get,
${\text{Perpendicular}} = 12{\text{ units}}$ and ${\text{Hypotenuse}} = 13{\text{ units}}$ .
From Pythagoras theorem we have,
${\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {{\text{Base}}} \right)^2} + {\left( {{\text{Perpendicular}}} \right)^2}$
Putting the values of hypotenuse and perpendicular we get,
${\left( {13} \right)^2} = {\left( {{\text{Base}}} \right)^2} + {\left( {12} \right)^2}$
$\Rightarrow {\left( {{\text{Base}}} \right)^2} = {\left( {13} \right)^2} - {\left( {12} \right)^2}$
$\Rightarrow {\left( {{\text{Base}}} \right)^2} = 169 - 144$
$\Rightarrow {\left( {{\text{Base}}} \right)^2} = 25$
$\Rightarrow {\text{Base}} = \sqrt {25}$
$\Rightarrow {\text{Base}} = 5{\text{ units}}$
Putting the values of base and hypotenuse in equation (ii) we get,
$\cos \theta = \dfrac{5}{{12}}$
From trigonometric identities we have,
$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
Putting the values of $\sin \theta$ and $\cos \theta$ we get,
$\tan \theta = \dfrac{{\left( {\dfrac{{12}}{{13}}} \right)}}{{\left( {\dfrac{5}{{12}}} \right)}}$
$\Rightarrow \tan \theta = \dfrac{{12}}{5}$
Now, we put the values of ${\text{sin}}\theta$ , $\cos \theta$ and $\tan \theta$ in the expression $\dfrac{{{{\sin }^2}\theta - {{\cos }^2}\theta }}{{2\sin \theta \cos \theta }} \times \dfrac{1}{{{{\tan }^2}\theta }}$ and we get,
$\dfrac{{{{\sin }^2}\theta - {{\cos }^2}\theta }}{{2\sin \theta \cos \theta }} \times \dfrac{1}{{{{\tan }^2}\theta }} = \dfrac{{{{\left( {\dfrac{{12}}{{13}}} \right)}^2} - {{\left( {\dfrac{5}{{13}}} \right)}^2}}}{{2 \times \left( {\dfrac{{12}}{{13}}} \right)\left( {\dfrac{5}{{13}}} \right)}} \times \dfrac{1}{{{{\left( {\dfrac{{12}}{5}} \right)}^2}}}$
$= \dfrac{{\left( {\dfrac{{144}}{{169}}} \right) - \left( {\dfrac{{25}}{{169}}} \right)}}{{2 \times \left( {\dfrac{{60}}{{169}}} \right)}} \times \dfrac{1}{{\left( {\dfrac{{144}}{{25}}} \right)}}$
$= \dfrac{{\left( {\dfrac{{144}}{{169}}} \right) - \left( {\dfrac{{25}}{{169}}} \right)}}{{\left( {\dfrac{{120}}{{169}}} \right)}} \times \dfrac{{25}}{{144}}$
$= \dfrac{{144 - 25}}{{120}} \times \dfrac{{25}}{{144}}$
$= \dfrac{{119}}{{120}} \times \dfrac{{25}}{{144}}$
$= \dfrac{{595}}{{3456}}$
Therefore, the value of $\dfrac{{{{\sin }^2}\theta - {{\cos }^2}\theta }}{{2\sin \theta \cos \theta }} \times \dfrac{1}{{{{\tan }^2}\theta }}$ is $\dfrac{{595}}{{3456}}$ .
So, the correct answer is “ $\dfrac{{595}}{{3456}}$ ”.

Note : There are three main functions in trigonometry, these are sine, cosine and tangent. There are three other trigonometric functions which can be written in terms of the main functions, these are cosecant which is inverse of sine, secant which is inverse of cosine and cotangent which is inverse of tangent. Also, while solving questions related to trigonometry, you should always remember the basic trigonometric identities.