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Question: How do you prove \(\dfrac{1}{{1 - \cos A}} + \dfrac{1}{{1 + \cos A}} = 2 + 2{\cot ^2}A\) ?...

How do you prove 11−cos⁡A+11+cos⁡A=2+2cot⁡2A\dfrac{1}{{1 - \cos A}} + \dfrac{1}{{1 + \cos A}} = 2 + 2{\cot ^2}A ?

Explanation

Solution

In this question, we are given one trigonometric function that is equal to some other trigonometric function. We don’t have to find the value of the “x” rather we have to prove that the trigonometric function given on the left-hand side is equal to the trigonometric function given on the right-hand side. We can prove this result by solving one of the sides of the given equation and try to make it equal to the other side.

Complete step by step answer:
We are given that –
⇒11−cos⁡A+11+cos⁡A=2+2cot⁡2A\Rightarrow \dfrac{1}{{1 - \cos A}} + \dfrac{1}{{1 + \cos A}} = 2 + 2{\cot ^2}A
Solving the left-hand side, we get –
⇒11−cos⁡A+11+cos⁡A ⇒1+cos⁡A+1−cos⁡A(1−cos⁡A)(1+cos⁡A) ⇒21−cos⁡2A       ∵(a−b)(a+b)=a2−b2  \Rightarrow \dfrac{1}{{1 - \cos A}} + \dfrac{1}{{1 + \cos A}} \\\ \Rightarrow \dfrac{{1 + \cos A + 1 - \cos A}}{{(1 - \cos A)(1 + \cos A)}} \\\ \Rightarrow \dfrac{2}{{1 - {{\cos }^2}A}}\,\,\,\,\,\,\,\because (a - b)(a + b) = {a^2} - {b^2} \\\
We know that
⇒sin⁡2x+cos⁡2x=1 ⇒1−cos⁡2x=sin⁡2x  \Rightarrow {\sin ^2}x + {\cos ^2}x = 1 \\\ \Rightarrow 1 - {\cos ^2}x = {\sin ^2}x \\\
Using this value in the obtained equation, we get –
⇒2sin⁡2A ⇒2cos⁡ec2A  \Rightarrow \dfrac{2}{{{{\sin }^2}A}} \\\ \Rightarrow 2\cos e{c^2}A \\\
We know that
⇒cos⁡ec2x−cot⁡2x=1 ⇒cos⁡ec2x=1+cot⁡2x  \Rightarrow \cos e{c^2}x - {\cot ^2}x = 1 \\\ \Rightarrow \cos e{c^2}x = 1 + {\cot ^2}x \\\
Using this identity in the obtained equation, we get –
⇒2(1+cot⁡2A) ⇒2+2cot⁡2A  \Rightarrow 2(1 + {\cot ^2}A) \\\ \Rightarrow 2 + 2{\cot ^2}A \\\
Now, the left-hand side has become equal to the right-hand side.
Hence proved that 11−cos⁡A+11+cos⁡A=2+2cot⁡2A\dfrac{1}{{1 - \cos A}} + \dfrac{1}{{1 + \cos A}} = 2 + 2{\cot ^2}A.

Note: Trigonometric functions are of six types, that is, sine function, cosine function, tangent function, cosecant function, secant function, and cotangent function. All these functions tell us the relation between two sides of the right-angled triangle and any one of the angles other than the right angle. All the trigonometric functions are related to each other by several identities, like in this question we have used the identity that the sum of the square of the sine and cosine of the same angle is equal to one, and the difference of the squares of the cosecant and cotangent of the same angle is equal to one.