Question

How do you prove \[1 + {\cot ^2}x = {\csc ^2}x\]?

Answer 1

Hint: Here, we will first take into consideration the identity of the sum of squares of the sine function and cosine function. Then we will divide the equation by the square of the sine function and use the trigonometric relation to simplify the equation. Using this we will prove the given identity.

Formula used:
The formulae used for solving this question are given by
(1) \[{\sin ^2}x + {\cos ^2}x = 1\]
(2) \[\dfrac{{\cos x}}{{\sin x}} = \cot x\]
(3) \[\dfrac{1}{{\sin x}} = {\mathop{\rm cosec}\nolimits} x\]

Complete step-by-step solution:
We know the trigonometric identity \[{\sin ^2}x + {\cos ^2}x = 1\].
Dividing both sides of the above equation by \[{\sin ^2}x\], we get
\[\dfrac{{{{\sin }^2}x + {{\cos }^2}x}}{{{{\sin }^2}x}} = \dfrac{1}{{{{\sin }^2}x}}\]
Splitting the fraction on the left hand side of the above equation into two fractions, we get
\[\dfrac{{{{\sin }^2}x}}{{{{\sin }^2}x}} + \dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}} = \dfrac{1}{{{{\sin }^2}x}}\]
\[ \Rightarrow 1 + {\left( {\dfrac{{\cos x}}{{\sin x}}} \right)^2} = {\left( {\dfrac{1}{{\sin x}}} \right)^2}\]……………………………\[\left( 1 \right)\]
Now, we know that
\[\dfrac{{\cos x}}{{\sin x}} = \cot x\]…………………………….\[\left( 2 \right)\]
\[\dfrac{1}{{\sin x}} = {\mathop{\rm cosec}\nolimits} x\]…………………………….\[\left( 3 \right)\]
Putting equation \[\left( 2 \right)\] and \[\left( 3 \right)\] in equation \[\left( 1 \right)\], we get
\[1 + {\left( {\cot x} \right)^2} = {\left( {{\mathop{\rm cosec}\nolimits} x} \right)^2}\]
Applying the exponent on the terms, we get
\[ \Rightarrow 1 + {\cot ^2}x = cose{c^2}x\]

Hence, the given identity \[1 + {\cot ^2}x = {\csc ^2}x\] is proved.

Note:
The identity \[{\sin ^2}x + {\cos ^2}x = 1\] is useful for deriving many other important trigonometric identities too. For example, by dividing the both sides of this identity by \[{\cos ^2}x\], we can prove another important identity which is written as \[1 + {\tan ^2}x = {\sec ^2}x\]. It all depends on the right hand side of the identity to be proved that with which trigonometric term, \[{\cos ^2}x\] or \[{\sin ^2}x\], the identity is to be divided. For example, for proving the trigonometric identity \[1 + {\cot ^2}x = {\csc ^2}x\], we divided the identity by the trigonometric term \[{\sin ^2}x\] since it is a reciprocal of \[{{\mathop{\rm cosec}\nolimits} ^2}x\].