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Question: How do you prove \[\dfrac{{{{\sin }^2}x}}{{1 - \cos x}} = 1 + \cos x\] ?...

How do you prove sin⁡2x1−cos⁡x=1+cos⁡x\dfrac{{{{\sin }^2}x}}{{1 - \cos x}} = 1 + \cos x ?

Explanation

Solution

Trigonometric functions are those functions that tell us the relation between the three sides of a right-angled triangle. We know that the Pythagoras relation between sine and cosine that is sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1 and also the algebraic identity a2−b2=(a+b)(a−b){a^2} - {b^2} = (a + b)(a - b). Using these identities we can solve this.

Complete step by step answer:
Given, sin⁡2x1−cos⁡x=1+cos⁡x\dfrac{{{{\sin }^2}x}}{{1 - \cos x}} = 1 + \cos x
We take the left hand side of the equation and then we show that it is equal to the right hand side of the equation.
LHS=sin⁡2x1−cos⁡xLHS = \dfrac{{{{\sin }^2}x}}{{1 - \cos x}} and RHS=1+cos⁡xRHS = 1 + \cos x
We have Pythagoras identity,
sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1
Rearranging we have,
sin⁡2x=1−cos⁡2x{\sin ^2}x = 1 - {\cos ^2}x
Substituting this value in the left hand side of the equation,
LHS=1−cos⁡2x1−cos⁡xLHS = \dfrac{{1 - {{\cos }^2}x}}{{1 - \cos x}}
This can be rewrite it as,
=12−cos⁡2x1−cos⁡x= \dfrac{{{1^2} - {{\cos }^2}x}}{{1 - \cos x}}
We know the algebraic identity a2−b2=(a+b)(a−b){a^2} - {b^2} = (a + b)(a - b). Using this the numerator term we have a=1a = 1and b=cos⁡xb = \cos x. Then above becomes,
(1−cos⁡x)(1+cos⁡x)(1−cos⁡x)\dfrac{{\left( {1 - \cos x} \right)\left( {1 + \cos x} \right)}}{{\left( {1 - \cos x} \right)}}
Cancelling the terms we have,
(1+cos⁡x)(1 + \cos x)
Thus we have LHS=1+cos⁡xLHS = 1 + \cos x
We have taken RHS=1+cos⁡xRHS = 1 + \cos x then we have
LHS=RHSLHS = RHS.
The left hand side of the equation is equal to the right hand side of the equation.

Hence we have proved sin⁡2x1−cos⁡x=1+cos⁡x\dfrac{{{{\sin }^2}x}}{{1 - \cos x}} = 1 + \cos x.

Note: Sine, cosine, tangent, cosecant, secant and cotangent are the six types of trigonometric functions; sine, cosine and tangent are the main functions while cosecant, secant and cotangent are the reciprocal of sine, cosine and tangent respectively. Remember A graph is divided into four quadrants, all the trigonometric functions are positive in the first quadrant, all the trigonometric functions are negative in the second quadrant except sine and cosine functions, tangent and cotangent are positive in the third quadrant while all others are negative and similarly all the trigonometric functions are negative in the fourth quadrant except cosine and secant.