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:
According to the given question, we need to convert the given equation in the form of sin .As the only value of sin is given so after converting into the form of sin with the help of trigonometry formula \[{\cos ^2}\theta \] = \[1 - {\sin ^2}\theta \]we just need to put the given value to evaluate the result.

Formula used:
Here, we use the formula of trigonometric functions that is \[{\sin ^2}\theta + {\cos ^2}\theta = 1\] which is further equals to \[{\cos ^2}\theta \] = \[1 - {\sin ^2}\theta \].

Complete step by step solution:
As, it is given \[\sin \theta = \dfrac{{12}}{{13}}\] and we have to find the value of \[\dfrac{{{{\sin }^2}\theta - {{\cos }^2}\theta }}{{2\sin \theta \cos \theta }} \times \dfrac{1}{{{{\tan }^2}\theta }}\] .
So, firstly we will convert tan in sin and cos form.
Here, we will get \[\dfrac{{{{\sin }^2}\theta - {{\cos }^2}\theta }}{{2\sin \theta \cos \theta }} \times \dfrac{1}{{\dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}}}\]
On simplifying we get, \[\dfrac{{{{\sin }^2}\theta - {{\cos }^2}\theta }}{{2\sin \theta \cos \theta }} \times \dfrac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }}\]
Now, using the identity we will replace \[{\cos ^2}\theta \] with \[1 - {\sin ^2}\theta \] it is derived from the identity \[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
We will first substitute the values in \[\dfrac{{{{\sin }^2}\theta - {{\cos }^2}\theta }}{{2\sin \theta \cos \theta }}\] we get,
\[ \Rightarrow \dfrac{{{{\sin }^2}\theta - (1 - {{\sin }^2}\theta )}}{{2\sin \theta \cos \theta }}\]
\[ \Rightarrow \dfrac{{{{\sin }^2}\theta - 1 + {{\sin }^2}\theta }}{{2\sin \theta \cos \theta }}\]
Also, \[ \Rightarrow \dfrac{{2{{\sin }^2}\theta - 1}}{{2\sin \theta \cos \theta }}\]
Cancelling 2 from both numerator and denominator we get,
\[ \Rightarrow \dfrac{{{{\sin }^2}\theta - \dfrac{1}{2}}}{{\sin \theta \cos \theta }}\]
Now, we will solve \[\dfrac{{{{\sin }^2}\theta - \dfrac{1}{2}}}{{\sin \theta \cos \theta }} \times \dfrac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }}\]
Also, cancelling \[\cos \theta \] from numerator and denominator.
\[ \Rightarrow \dfrac{{{{\sin }^2}\theta - \dfrac{1}{2}}}{{\sin \theta }} \times \dfrac{{\cos \theta }}{{{{\sin }^2}\theta }}\]
Substituting the value of \[\cos \theta \] as \[\sqrt {1 - {{\sin }^2}\theta } \] and also multiplying denominators with each other.
We get,
\[ \Rightarrow \dfrac{{{{\sin }^2}\theta - \dfrac{1}{2}}}{{{{\sin }^3}\theta }} \times \sqrt {1 - {{\sin }^2}\theta } \]
As, above equation is converted into sin form so we will substitute the value of \[\sin \theta = \dfrac{{12}}{{13}}\]
On putting the values we get,
\[ \Rightarrow \dfrac{{{{\left( {\dfrac{{12}}{{13}}} \right)}^2} - \dfrac{1}{2}}}{{{{\left( {\dfrac{{12}}{{13}}} \right)}^3}}} \times \sqrt {1 - {{\left( {\dfrac{{12}}{{13}}} \right)}^2}} \]
As we know squares of 12 and 13 are 144 and 169 respectively. And the cube of 12 is 1728 and 13 is 2197.
So, on substituting the values of squares and cubes we get,
 \[ \Rightarrow \dfrac{{\dfrac{{144}}{{169}} - \dfrac{1}{2}}}{{\dfrac{{1728}}{{2197}}}} \times \sqrt {1 - \dfrac{{144}}{{169}}} \]
By taking the L.C.M we get,
\[ \Rightarrow \dfrac{{\dfrac{{144 \times 2 - 1 \times 169}}{{169 \times 2}}}}{{\dfrac{{1728}}{{2197}}}} \times \sqrt {\dfrac{{169 - 144}}{{169}}} \]
On simplifying the multiplication:
\[ \Rightarrow \dfrac{{\dfrac{{288 - 169}}{{338}}}}{{\dfrac{{1728}}{{2197}}}} \times \sqrt {\dfrac{{169 - 144}}{{169}}} \]
Also, \[ \Rightarrow \dfrac{{\dfrac{{119}}{{338}}}}{{\dfrac{{1728}}{{2197}}}} \times \sqrt {\dfrac{{25}}{{169}}} \]
As we know the square root of 25 and 169 are 5 and 13. Thus, we can calculate using the prime factorization method.
After substituting we get,
\[ \Rightarrow \dfrac{{119}}{{338}} \times \dfrac{{2197}}{{1728}} \times \dfrac{5}{{13}}\]
After simplification of the above equation:
\[ \Rightarrow \dfrac{{119}}{2} \times \dfrac{1}{{1728}} \times \dfrac{5}{1}\]
On Multiply the denominator and numerator we get,
\[ \Rightarrow \dfrac{{595}}{{3456}}\]

Hence, the value of \[\dfrac{{{{\sin }^2}\theta - {{\cos }^2}\theta }}{{2\sin \theta \cos \theta }} \times \dfrac{1}{{{{\tan }^2}\theta }}\]\[ = \dfrac{{595}}{{3456}}\]

Note:
To solve these types of questions, we have one alternative method also that is with the help of sin value calculate the value of cos and tan by using Pythagoras theorem and then put the calculated values to find the answer . So, it is concluded that there can be different ways to solve the given question.