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Question: Prove that: \[\dfrac{{1 + \cos A}}{{\sin A}} = \dfrac{{\sin A}}{{1 - \cos A}}\]...

Prove that: 1+cos⁡Asin⁡A=sin⁡A1−cos⁡A\dfrac{{1 + \cos A}}{{\sin A}} = \dfrac{{\sin A}}{{1 - \cos A}}

Explanation

Solution

Hint : Here we need to prove so, we will take the LHS and try to get the RHS from it. We will first multiply both numerator as well as denominator by (1−cos⁡A)\left( {1 - \cos A} \right), and then we will simplify the resulting expression using the trigonometric formulas.
Formula used:
sin⁡2A+cos⁡2A=1{\sin ^2}A + {\cos ^2}A = 1
(a+b)(a−b)=a2−b2\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}

Complete step by step solution:
We have the LHS;
=1+cos⁡Asin⁡A= \dfrac{{1 + \cos A}}{{\sin A}}
Now we can see that in the numerator of the RHS we have sin⁡A\sin A, so, we will try to get sin⁡A\sin A in the numerator of the above fraction and for this we will multiply both the numerator and denominator by (1−cos⁡A)\left( {1 - \cos A} \right), because we know, (1−cos⁡A)(1+cos⁡A)=sin⁡2A\left( {1 - \cos A} \right)\left( {1 + \cos A} \right) = {\sin ^2}A.
So, multiplying numerator and denominator by (1−cos⁡A)\left( {1 - \cos A} \right),
=1+cos⁡Asin⁡A×1−cos⁡A1−cos⁡A= \dfrac{{1 + \cos A}}{{\sin A}} \times \dfrac{{1 - \cos A}}{{1 - \cos A}}
Now we will simplify the numerator using the formula that, (a+b)(a−b)=a2−b2\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}.
So, we get,
=1−cos⁡2Asin⁡A(1−cos⁡A)= \dfrac{{1 - {{\cos }^2}A}}{{\sin A\left( {1 - \cos A} \right)}}
Now we know the formula that,
sin⁡2A+cos⁡2A=1{\sin ^2}A + {\cos ^2}A = 1
⇒1−cos⁡2A=sin⁡2A\Rightarrow 1 - {\cos ^2}A = {\sin ^2}A
Now, we will use this formula in the numerator. So, we get;
=sin⁡2Asin⁡A(1−cos⁡A)= \dfrac{{{{\sin }^2}A}}{{\sin A\left( {1 - \cos A} \right)}}
Now we will cancel the sin⁡A\sin A term from the numerator and denominator. So, we get;
=sin⁡A1−cos⁡A= \dfrac{{\sin A}}{{1 - \cos A}}
Which is the RHS term.

Note : Here in order to solve this question one can also think of starting with the RHS term and getting the LHS term from it. But that will either get lengthy or one might not be able to get the required results. Also, in some other questions it is possible that if we start with the RHS term we will get the LHS term more easily than if we start with the LHS term and try to get the RHS term. So, it is very important to decide in the beginning how we choose to prove the given question.