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Question: Prove that: \( \left( {\dfrac{{\sin \theta + \cos \theta }}{{\sin \theta - \cos \theta }}} \right...

Prove that:
(sin⁡θ+cos⁡θsin⁡θ−cos⁡θ)+(sin⁡θ−cos⁡θsin⁡θ+cos⁡θ)=22sin⁡2θ−1\left( {\dfrac{{\sin \theta + \cos \theta }}{{\sin \theta - \cos \theta }}} \right) + \left( {\dfrac{{\sin \theta - \cos \theta }}{{\sin \theta + \cos \theta }}} \right) = \dfrac{2}{{2{{\sin }^2}\theta - 1}}

Explanation

Solution

Hint : Cross multiply the LHS and then use the expansion formula of square to expand it. Then use the trigonometric identity of sin⁡θ\sin \theta and cos⁡θ\cos \theta to simplify the expression. Keep on simplifying it until it turns equal to RHS.

Complete step-by-step answer :
The equation given in the question is
⇒(sin⁡θ+cos⁡θsin⁡θ−cos⁡θ)+(sin⁡θ−cos⁡θsin⁡θ+cos⁡θ)=22sin⁡2θ−1\Rightarrow \left( {\dfrac{{\sin \theta + \cos \theta }}{{\sin \theta - \cos \theta }}} \right) + \left( {\dfrac{{\sin \theta - \cos \theta }}{{\sin \theta + \cos \theta }}} \right) = \dfrac{2}{{2{{\sin }^2}\theta - 1}}
Then, LHS
=(sin⁡θ+cos⁡θsin⁡θ−cos⁡θ)+(sin⁡θ−cos⁡θsin⁡θ+cos⁡θ)= \left( {\dfrac{{\sin \theta + \cos \theta }}{{\sin \theta - \cos \theta }}} \right) + \left( {\dfrac{{\sin \theta - \cos \theta }}{{\sin \theta + \cos \theta }}} \right)
By cross multiplying, we get
=(sin⁡θ+cos⁡θ)2+(sin⁡θ−cos⁡θ)2(sin⁡θ−cos⁡θ)(sin⁡θ+cos⁡θ)= \dfrac{{{{\left( {\sin \theta + \cos \theta } \right)}^2} + {{\left( {\sin \theta - \cos \theta } \right)}^2}}}{{\left( {\sin \theta - \cos \theta } \right)\left( {\sin \theta + \cos \theta } \right)}}
Now, by using the formulae
(a+b)=a2+b2+2ab,(a−b)2=a2+b2−2ab(a + b) = {a^2} + {b^2} + 2ab,{(a - b)^2} = {a^2} + {b^2} - 2ab and (a−b)(a+b)=a2−b2(a - b)(a + b) = {a^2} - {b^2}
We can write the above expression as
=sin⁡2θ+cos⁡2θ+2sin⁡θcos⁡θ+sin⁡2θ+cos⁡2θ−2sin⁡θcos⁡θsin⁡2θ−cos⁡2θ= \dfrac{{{{\sin }^2}\theta + {{\cos }^2}\theta + 2\sin \theta \cos \theta + {{\sin }^2}\theta + {{\cos }^2}\theta - 2\sin \theta \cos \theta }}{{{{\sin }^2}\theta - {{\cos }^2}\theta }}
Now we know that,
sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1
By using this identity and cancelling the terms with opposite signs, we can write
=1+1sin⁡2θ−(1−sin⁡2θ)= \dfrac{{1 + 1}}{{{{\sin }^2}\theta - (1 - {{\sin }^2}\theta )}}
On simplifying it, we get
=2sin⁡2θ−1+sin⁡2θ= \dfrac{2}{{{{\sin }^2}\theta - 1 + {{\sin }^2}\theta }}
=22sin⁡2θ−1= \dfrac{2}{{2{{\sin }^2}\theta - 1}}
= RHS
Hence it is proved that,
(sin⁡θ+cos⁡θsin⁡θ−cos⁡θ)+(sin⁡θ−cos⁡θsin⁡θ+cos⁡θ)=22sin⁡2θ−1\left( {\dfrac{{\sin \theta + \cos \theta }}{{\sin \theta - \cos \theta }}} \right) + \left( {\dfrac{{\sin \theta - \cos \theta }}{{\sin \theta + \cos \theta }}} \right) = \dfrac{2}{{2{{\sin }^2}\theta - 1}}

Note : It should click you as soon as you see the question that you need to cross multiply it. Because in LHS, the term in the numerator of the first term is equal to the term in the denominator of the second term. So it was very clear that if we cross multiply, then it can be expanded using the formula of (a+b)2{(a + b)^2} which further can be simplified using trigonometric identities.