Question

How do you use the fundamental identities to simplify $\dfrac{{1 - {{\sin }^2}x}}{{{{\csc }^2}x - 1}}$ ?

Answer 1

Hint : The identity that we will use in this question is that the sum of the squares of sine function and cosine function is equal to one, and the other identity used states that the difference of the squares of the cosecant and cotangent function is one. Using these identities we can simplify the fraction, and then we will convert all the trigonometric ratios in the form of the sine and cosine functions. The trigonometric terms are now simplified, now we will simplify the fraction by canceling out the common terms present in the numerator and the denominator.

Complete step-by-step answer :
In this question, we are given
Complete step by step answer:
We are given $\dfrac{{1 - {{\sin }^2}x}}{{{{\csc }^2}x - 1}}$
We know that,
${\sin ^2}x + {\cos ^2}x = 1 \\\ \Rightarrow {\cos ^2}x = 1 - {\sin ^2}x \;$
And
${\csc ^2}x - {\cot ^2}x = 1 \\\ \Rightarrow {\csc ^2}x - 1 = {\cot ^2}x \;$
Using the above two values in the given expression, we get –
$\dfrac{{1 - {{\sin }^2}x}}{{{{\csc }^2}x - 1}} = \dfrac{{{{\cos }^2}x}}{{{{\cot }^2}x}}$
Now, $\cot x = \dfrac{{\cos x}}{{\sin x}}$
So,
$\dfrac{{1 - {{\sin }^2}x}}{{{{\csc }^2}x - 1}} = \dfrac{{{{\cos }^2}x}}{{\dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}}}} = {\cos ^2}x \times \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} \\\ \Rightarrow \dfrac{{1 - {{\sin }^2}x}}{{{{\csc }^2}x - 1}} = {\sin ^2}x \;$
Hence, the simplified form of $\dfrac{{1 - {{\sin }^2}x}}{{{{\csc }^2}x - 1}}$ is ${\sin ^2}x$
So, the correct answer is “ ${\sin ^2}x$ ”.

Note : We can solve this question by one more method as shown below –
We know that $\csc x = \dfrac{1}{{\sin x}}$
Using this value in the above equation, we get –
$\dfrac{{1 - {{\sin }^2}x}}{{{{\csc }^2}x - 1}} = \dfrac{{1 - {{\sin }^2}x}}{{\dfrac{1}{{{{\sin }^2}x}} - 1}} = \dfrac{{1 - {{\sin }^2}x}}{{\dfrac{{1 - {{\sin }^2}x}}{{{{\sin }^2}x}}}} = 1 - {\sin ^2}x \times \dfrac{{{{\sin }^2}x}}{{1 - {{\sin }^2}x}}$
Now the term $1 - {\sin ^2}x$ is common in both the numerator and the denominator, so we cancel it out as follows –
$\Rightarrow \dfrac{{1 - {{\sin }^2}x}}{{{{\csc }^2}x - 1}} = {\sin ^2}x$