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How do you prove the identity \[\dfrac{1}{{1 - \cos x}} + \dfrac{1}{{1 + \cos x}} = 2{\csc ^2}x\]?

Last updated date: 17th Jul 2024
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Hint: Here we will first take into consideration the expression on the left-hand side of the equation. We will take the LCM of the fractions and simplify it further using the algebraic identity. Then we will apply the trigonometric reciprocal function to prove the given identity.

Complete step by step solution:
We have to prove that \[\dfrac{1}{{1 - \cos x}} + \dfrac{1}{{1 + \cos x}} = 2{\csc ^2}x\]
Taking Left hand side value we get,
LHS \[ = \dfrac{1}{{1 - \cos x}} + \dfrac{1}{{1 + \cos x}}\]
Now we will take the L.CM of the above terms, so we get
\[ \Rightarrow \dfrac{1}{{1 - \cos x}} + \dfrac{1}{{1 + \cos x}} = \dfrac{{\left( {1 + \cos x} \right) + \left( {1 - \cos x} \right)}}{{\left( {1 + \cos x} \right)\left( {1 - \cos x} \right)}}\]
On solving the denominator by using the algebraic identity \[\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}\], we get,
\[ \Rightarrow \dfrac{1}{{1 - \cos x}} + \dfrac{1}{{1 + \cos x}} = \dfrac{{1 + \cos x + 1 - \cos x}}{{1 - {{\cos }^2}x}}\]
Adding and subtracting the like terms, we get
\[ \Rightarrow \dfrac{1}{{1 - \cos x}} + \dfrac{1}{{1 + \cos x}} = \dfrac{2}{{1 - {{\cos }^2}x}}\]……\[\left( 1 \right)\]
We know the trigonometric identity, \[{\sin ^2}x + {\cos ^2}x = 1\]. Rewriting the identity, we get
 \[{\sin ^2}x = 1 - {\cos ^2}x\]
Substituting the above value in equation \[\left( 1 \right)\], we get
\[ \Rightarrow \dfrac{1}{{1 - \cos x}} + \dfrac{1}{{1 + \cos x}} = \dfrac{2}{{{{\sin }^2}x}}\]
Now using the reciprocal function \[\csc x = \dfrac{1}{{\sin x}}\], we get
\[ \Rightarrow \dfrac{1}{{1 - \cos x}} + \dfrac{1}{{1 + \cos x}} = 2{\csc ^2}x\]
We can see that LHS \[ = \] RHS.
Hence, we have proved the identity.

Additional information:
Trigonometry is a branch of mathematics that deals with specific functions of angles and also their application in calculations. The commonly used six types of trigonometry functions are sine, cosine, tangent, cotangent, secant, and cosecant. Where cosecant is reciprocal of sine secant is reciprocal of cosine and cotangent is reciprocal of a tangent. Trigonometric identities are equalities that have trigonometric functions and which are true for all occurring variables and both sides of the equation are given. We use these identities whenever we have to solve the expressions having trigonometric functions in them.

We can also solve this question using an alternate method. We can approach the solution from the right-hand side of the equation.
RHS \[ = 2{\csc ^2}x\]
Now using the reciprocal trigonometric function, \[\csc x = \dfrac{1}{{\sin x}}\], we get
\[ \Rightarrow 2{\csc ^2}x = \dfrac{2}{{{{\sin }^2}x}}\]
Now substituting \[{\sin ^2}x = 1 - {\cos ^2}x\], we get
\[ \Rightarrow 2{\csc ^2}x = \dfrac{2}{{1 - {{\cos }^2}x}}\]
We can rewrite the above equation as
\[ \Rightarrow 2{\csc ^2}x = \dfrac{2}{{{1^2} - {{\cos }^2}x}}\]
Using the algebraic identity \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\], we get
\[ \Rightarrow 2{\csc ^2}x = \dfrac{2}{{\left( {1 - \cos x} \right)\left( {1 + \cos x} \right)}}\]
We can write the above equation as
\[ \Rightarrow 2{\csc ^2}x = \dfrac{1}{{\left( {1 - \cos x} \right)}} + \dfrac{1}{{\left( {1 + \cos x} \right)}}\]
Hence proved.