Question

How do you identify ${\cot ^2}x({\sin ^2}x)$ ?

Answer 1

Hint : First we will evaluate the right-hand of the equation and then further the left-hand side of the equation. We will use the identity ${\sin ^2}x + {\cos ^2}x = 1$ if required. Then we will try to factorise and simplify the terms so that the left-hand side matches the right-hand side.

Complete step-by-step answer :
We will start off by solving the right-hand side of the equation. Here, we will be first making the denominator of the left-hand side the same.
Here, we are using the trigonometric identity,
$\cot x = \,\dfrac{{\cos x}}{{\sin x}}$
Hence, the expression can be written as,
$= {\cot ^2}x({\sin ^2}x) \\\ \, = {\left( {\dfrac{{\cos x}}{{\sin x}}} \right)^2}({\sin ^2}x) \\\ = \left( {\dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}}} \right)({\sin ^2}x) \;$
Now we cancel out all the same terms.
$= \left( {\dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}}} \right)({\sin ^2}x) \\\ = {\cos ^2}x \;$
Hence, the value of the expression ${\cot ^2}x({\sin ^2}x)$ is ${\cos ^2}x$ .
So, the correct answer is “ ${\cos ^2}x$ ”.

Note : While choosing the side to solve, always choose the side where you can directly apply the trigonometric identities. Also, remember the trigonometric identities ${\sin ^2}x + {\cos ^2}x = 1$ and $\cos 2x = 2{\cos ^2}x - 1$ . While opening the brackets make sure you are opening the brackets properly with their respective signs. Also remember that $\cot x = \,\dfrac{{\cos x}}{{\sin x}}$ .
While applying the double angle identities, first choose the identity according to the terms you have then choose the terms from the expression involving which you are using the double angle identities. While modifying any identity make sure that when you back trace the identity, you get the same original identity.