Question

How do you factor the expression and use the fundamental identities to simplify \[{\sin ^2}x{\csc ^2}x - {\sin ^2}x\] ?

Answer 1

Hint: We are given the identity \[{\sin ^2}x{\csc ^2}x - {\sin ^2}x\], we will first take \[{\sin ^2}x\] common from the given expression to write it in form of factor as \[{\sin ^2}x\left( {{{\csc }^2}x - 1} \right)\]. Then we will use the trigonometric identity \[{\csc ^2}x - {\cot ^2}x = 1\] to rewrite the expression. At last, we will cancel the common terms to obtain the simplified form.

Complete step by step answer:
We are given the expression, \[{\sin ^2}x{\csc ^2}x - {\sin ^2}x\].
Taking \[{\sin ^2}x\] common, we get
\[ \Rightarrow {\sin ^2}x{\csc ^2}x - {\sin ^2}x = {\sin ^2}x\left( {{{\csc }^2}x - 1} \right)\]
As we know from the trigonometric identity that \[{\csc ^2}x - {\cot ^2}x = 1\] i.e., \[{\csc ^2}x - 1 = {\cot ^2}x\].
Using this we get can write the given expression as,
\[ \Rightarrow {\sin ^2}x{\csc ^2}x - {\sin ^2}x = {\sin ^2}x\left( {{{\cot }^2}x} \right)\]
As we know that \[\cot x = \dfrac{{\cos x}}{{\sin x}}\]. Now, we will convert the given equation into \[\sin x\] and \[\cos x\].

So, we get
\[ \Rightarrow {\sin ^2}x{\csc ^2}x - {\sin ^2}x = {\sin ^2}x \times {\left( {\dfrac{{\cos x}}{{\sin x}}} \right)^2}\]
On simplification, we get
\[ \Rightarrow {\sin ^2}x{\csc ^2}x - {\sin ^2}x = {\sin ^2}x \times \dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}}\]
On cancelling the common terms from the numerator and the denominator, we get
\[ \therefore {\sin ^2}x{\csc ^2}x - {\sin ^2}x = {\cos ^2}x\]

Therefore, by simplifying \[{\sin ^2}x{\csc ^2}x - {\sin ^2}x\], we get \[{\cos ^2}x\].

Additional information: Trigonometric functions are real functions which relate an angle of a right-angled triangle to ratios of two sides. Trigonometric formulas are considered only for right angled triangles. We have three sides namely hypotenuse, perpendicular and base in a right-angled triangle. Hypotenuse is the longest side, side opposite to the angle is perpendicular and base is the side where both hypotenuse and perpendicular rests.

Note: Here, if it was not mentioned in the question that we have to factor the expression and then use the fundamental identities to simplify then we may solve this problem by another method.
Given expression is \[{\sin ^2}x{\csc ^2}x - {\sin ^2}x\].
As we know that \[\csc x = \dfrac{1}{{\sin x}}\]. Using this, we get the expression as
\[ \Rightarrow {\sin ^2}x{\csc ^2}x - {\sin ^2}x = \left( {{{\sin }^2}x \times \dfrac{1}{{{{\sin }^2}x}}} \right) - {\sin ^2}x\]
On cancelling the common terms from the numerator and the denominator, we get
\[ \Rightarrow {\sin ^2}x{\csc ^2}x - {\sin ^2}x = 1 - {\sin ^2}x\]
As we know from the trigonometric identity that \[{\sin ^2}x + {\cos ^2}x = 1\] i.e., \[1 - {\sin ^2}x = {\cos ^2}x\]. Using this, we can write
\[ \Rightarrow {\sin ^2}x{\csc ^2}x - {\sin ^2}x = {\cos ^2}x\]
Therefore, by simplifying \[{\sin ^2}x{\csc ^2}x - {\sin ^2}x\], we get \[{\cos ^2}x\].