Question

If \[\sin \theta = \dfrac{{12}}{{13}}\], find the value of \[\dfrac{{si{n^2}\theta - co{s^2}\theta }}{{2 \sin \theta \cos \theta }} \times \dfrac{1}{{{{\tan }^2}\theta }}\]?

Answer 1

Hint:
We have given the value of $\operatorname{Sin} \theta $ and we have to calculate the value of a trigonometric expression. The trigonometric expression contains the trigonometric function$\operatorname{Cos} \theta $,$\operatorname{Sin} \theta $, and $tan\theta $. Firstly we have to calculate the value of these trigonometric functions. These trigonometric functions can be calculated by the relation of sides of the right angle triangle with the function. We have given $\operatorname{Sin} \theta $,and $\operatorname{Sin} \theta $ is related with ‘perpendicular’ and ‘hypotenuse’ of the right angle triangle. So with the help of these We can calculate the value of ’Base’. This will help us in calculating the trigonometric function.

Complete step by step solution:
We have given that \[sin\theta = \dfrac{{12}}{{13}}\]
We have to calculate the \[\dfrac{{si{n^2}\theta - co{s^2}\theta }}{{2sin\theta cos\theta }} \times \dfrac{1}{{{{\tan }^2}\theta }}\]
Since \[sin\theta = \dfrac{{12}}{{13}}\]
We first calculate the value of $\operatorname{Cos} \theta $ and $tan\theta $.
We know that $\operatorname{Sin} \theta $ is related with the sides of right angle triangle as $\operatorname{Sin} \theta = \dfrac{{Perpendicular}}{{Hypotenuse}}$
Therefore $\operatorname{Sin} \theta = \dfrac{{Perpendicular}}{{Hypotenuse}}$\[ = \dfrac{{12}}{{13}}\]
\[Perpendicular{\text{ }} = {\text{ }}12k\], \[Hypotenuse = 13k\]
Here ‘K’ is a constant number which is common in both. While taking ratio they cancel each other
By Pythagorus theorem
\[{\left( {Hypotenuse} \right)^2} = {\left( {Perpendicular} \right)^2} + {\left( {Base} \right)^2}\]
\[\Rightarrow {\left( {13k} \right)^2} = {\left( {12k} \right)^2} + {\left( {Base} \right)^2}\]
\[Trigonometric Ratios169{k^2} = 144{k^2} + {\left( {Base} \right)^2}\]
\[Trigonometric Ratios169{k^2} - 144{k^2} = {\left( {Base} \right)^2}\]
\[Trigonometric Ratios25{k^2} = {\left( {Base} \right)^2}\]
\[Base = 5k\]

Now we have to calculate the value of $\operatorname{Cos} \theta $ and $tan\theta $
$\operatorname{Cos} \theta $ is related with sides of triangle as
$\operatorname{Cos} \theta $$ = \dfrac{B}{H}$$ = \dfrac{{5k}}{{13k}}$=$\dfrac{5}{{13}}$
$tan\theta $ is related with the sides of triangle as

$tan\theta $=$\dfrac{p}{b} = \dfrac{{12k}}{{5k}} = \dfrac{{12}}{5}$
Now, We calculate the value of trigonometric expression
 \[\dfrac{{si{n^2}\theta - co{s^2}\theta }}{{2sin\theta cos\theta }} \times \dfrac{1}{{{{\tan }^2}\theta }}\]\[ = \dfrac{{595}}{{3456}}\]
\[ = \dfrac{{{{\left( {\dfrac{{12}}{{13}}} \right)}^2} - {{\left( {\dfrac{5}{{13}}} \right)}^2}}}{{2 \times \dfrac{{12}}{{13}} \times \dfrac{5}{{13}}}} \times \dfrac{1}{{^{^{{{\left( {\dfrac{{12}}{5}} \right)}^2}}}}}\]
\[ = \dfrac{{{{\left( {\dfrac{{144}}{{169}}} \right)}^{}} - {{\left( {\dfrac{{25}}{{169}}} \right)}^{}}}}{{\dfrac{{120}}{{169}}}} \times \dfrac{1}{{\left( {\dfrac{{144}}{{25}}} \right)}}\]
\[ = \dfrac{{119}}{{120}} \times \dfrac{{25}}{{144}}\]
\[ = \dfrac{{595}}{{3456}}\]

So the value of \[\dfrac{{si{n^2}\theta - co{s^2}\theta }}{{2sin\theta cos\theta }} \times \dfrac{1}{{{{\tan }^2}\theta }}\]\[ = \dfrac{{595}}{{3456}}\]

Note:
Trigonometry is the branch of mathematics which studies the relationships between side lengths and angles of triangles. We have six trigonometric functions whose values at different angles can be calculated with the help of these relations.
Trigonometric function: The trigonometric functions are the real functions which relate the angle of right angles triangle to ratios of two side lengths.