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Question: Prove the following inverse trigonometric equation: \({{\cot }^{-1}}\left( \dfrac{\sqrt{1+\sin x}+...

Prove the following inverse trigonometric equation:
cot⁡−1(1+sin⁡x+1−sin⁡x1+sin⁡x−1−sin⁡x)=x2;x∈(0,π4).{{\cot }^{-1}}\left( \dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}} \right)=\dfrac{x}{2};x\in \left( 0,\dfrac{\pi }{4} \right).

Explanation

Solution

Hint: To solve expressions with irrational value in the denominator , first rationalise the denominator by multiplying it with its conjugate.

First we will consider the left hand side of the equation. We are given LHS=cot⁡−1(1+sin⁡x+1−sin⁡x1+sin⁡x−1−sin⁡x)LHS={{\cot }^{-1}}\left( \dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}} \right).
Now, we can see that the denominator is irrational . So , we need to rationalise it first. We can do this by multiplying and dividing by its conjugate , i.e. 1+sin⁡x+1−sin⁡x\sqrt{1+\sin x}+\sqrt{1-\sin x} .
So , we have LHS=cot⁡−1((1+sin⁡x+1−sin⁡x)1+sin⁡x−1−sin⁡x×1+sin⁡x+1−sin⁡x1+sin⁡x+1−sin⁡x)LHS={{\cot }^{-1}}\left( \dfrac{\left( \sqrt{1+\sin x}+\sqrt{1-\sin x} \right)}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\times \dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}+\sqrt{1-\sin x}} \right)
=cot⁡−1((1+sin⁡x+1−sin⁡x)2(1+sin⁡x−1−sin⁡x)(1+sin⁡x+1−sin⁡x))={{\cot }^{-1}}\left( \dfrac{{{\left( \sqrt{1+\sin x}+\sqrt{1-\sin x} \right)}^{2}}}{(\sqrt{1+\sin x}-\sqrt{1-\sin x})(\sqrt{1+\sin x}+\sqrt{1-\sin x})} \right)
Now, we know (a+b)2=a2+b2+2ab{{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab. Using this identity in the numerator, we get
=cot⁡−1(1+sin⁡x+1−sin⁡x+2(1+sin⁡x)(1−sin⁡x)(1+sin⁡x−1−sin⁡x)(1+sin⁡x+1−sin⁡x))={{\cot }^{-1}}\left( \dfrac{1+\sin x+1-\sin x+2\sqrt{\left( 1+\sin x \right)\left( 1-\sin x \right)}}{(\sqrt{1+\sin x}-\sqrt{1-\sin x})(\sqrt{1+\sin x}+\sqrt{1-\sin x})} \right)
Now , we know (a+b)(a−b)=a2−b2(a+b)(a-b)={{a}^{2}}-{{b}^{2}}. Using this identity in the denominator , we get
=cot⁡−1(2+2(1+sin⁡x)(1−sin⁡x)(1+sin⁡x)−(1−sin⁡x))={{\cot }^{-1}}\left( \dfrac{2+2\sqrt{\left( 1+\sin x \right)\left( 1-\sin x \right)}}{(1+\sin x)-(1-\sin x)} \right)
=cot⁡−1(2+21−sin⁡2x2sin⁡x)={{\cot }^{-1}}\left( \dfrac{2+2\sqrt{1-{{\sin }^{2}}x}}{2\sin x} \right)
Now, we know cos⁡2x+sin⁡2x=1⇒1−sin⁡2=cos⁡2x{{\cos }^{2}}x+{{\sin }^{2}}x=1\Rightarrow 1-{{\sin }^{2}}={{\cos }^{2}}x
Now, we will substitute 1−sin⁡2=cos⁡2x1-{{\sin }^{2}}={{\cos }^{2}}x in the numerator of the L.H.S .
On substituting 1−sin⁡2=cos⁡2x1-{{\sin }^{2}}={{\cos }^{2}}x in the numerator of the L.H.S , we get
L.H.S=cot⁡−1(2+2cos⁡x2sin⁡x)={{\cot }^{-1}}\left( \dfrac{2+2\cos x}{2\sin x} \right)
Now,
We know, 1+cos⁡θ=2cos⁡2θ21+\cos \theta =2{{\cos }^{2}}\dfrac{\theta }{2}
And sin⁡θ=2sin⁡θ2cos⁡θ2\sin \theta =2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}
So, we will substitute 1+cos⁡θ=2cos⁡2θ21+\cos \theta =2{{\cos }^{2}}\dfrac{\theta }{2} in the numerator of the L.H.S and sin⁡θ=2sin⁡θ2cos⁡θ2\sin \theta =2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}in the denominator of the L.H.S.
On substituting 1+cos⁡θ=2cos⁡2θ21+\cos \theta =2{{\cos }^{2}}\dfrac{\theta }{2} and sin⁡θ=2sin⁡θ2cos⁡θ2\sin \theta =2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2} in the left hand side of the equation, we get
L.H.S =cot⁡−1(2cos⁡2x22sin⁡x2cos⁡x2)={{\cot }^{-1}}\left( \dfrac{2{{\cos }^{2}}\dfrac{x}{2}}{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}} \right)
=cot⁡−1(cot⁡x2)={{\cot }^{-1}}\left( \cot \dfrac{x}{2} \right)
=x2=\dfrac{x}{2}
Now, we will consider the right hand side of the equation.
The value given in the right hand side of the equation is equal to x2\dfrac{x}{2}.
Now, we can see that the value obtained by solving the expression in the left hand side of the equation and the value given in the right hand side of the equation are equal.
So, L.H.S = R.H.S
Hence proved.

Note: Remember that conjugate of a+b\sqrt{a}+\sqrt{b} is a−b\sqrt{a}-\sqrt{b} and not (−a−b)\left( -\sqrt{a}-\sqrt{b} \right). Students generally make this mistake and get their answer wrong.