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More # Prove the following inverse trigonometric equation:${{\cot }^{-1}}\left( \dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}} \right)=\dfrac{x}{2};x\in \left( 0,\dfrac{\pi }{4} \right).$

Last updated date: 24th Mar 2023
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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 $LHS={{\cot }^{-1}}\left( \dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}} \right)$.
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. $\sqrt{1+\sin x}+\sqrt{1-\sin x}$ .
So , we have $LHS={{\cot }^{-1}}\left( \dfrac{\left( \sqrt{1+\sin x}+\sqrt{1-\sin x} \right)}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\times \dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}+\sqrt{1-\sin x}} \right)$
$={{\cot }^{-1}}\left( \dfrac{{{\left( \sqrt{1+\sin x}+\sqrt{1-\sin x} \right)}^{2}}}{(\sqrt{1+\sin x}-\sqrt{1-\sin x})(\sqrt{1+\sin x}+\sqrt{1-\sin x})} \right)$
Now, we know ${{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$. Using this identity in the numerator, we get
$={{\cot }^{-1}}\left( \dfrac{1+\sin x+1-\sin x+2\sqrt{\left( 1+\sin x \right)\left( 1-\sin x \right)}}{(\sqrt{1+\sin x}-\sqrt{1-\sin x})(\sqrt{1+\sin x}+\sqrt{1-\sin x})} \right)$
Now , we know $(a+b)(a-b)={{a}^{2}}-{{b}^{2}}$. Using this identity in the denominator , we get
$={{\cot }^{-1}}\left( \dfrac{2+2\sqrt{\left( 1+\sin x \right)\left( 1-\sin x \right)}}{(1+\sin x)-(1-\sin x)} \right)$
$={{\cot }^{-1}}\left( \dfrac{2+2\sqrt{1-{{\sin }^{2}}x}}{2\sin x} \right)$
Now, we know ${{\cos }^{2}}x+{{\sin }^{2}}x=1\Rightarrow 1-{{\sin }^{2}}={{\cos }^{2}}x$
Now, we will substitute $1-{{\sin }^{2}}={{\cos }^{2}}x$ in the numerator of the L.H.S .
On substituting $1-{{\sin }^{2}}={{\cos }^{2}}x$ in the numerator of the L.H.S , we get
L.H.S$={{\cot }^{-1}}\left( \dfrac{2+2\cos x}{2\sin x} \right)$
Now,
We know, $1+\cos \theta =2{{\cos }^{2}}\dfrac{\theta }{2}$
And $\sin \theta =2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$
So, we will substitute $1+\cos \theta =2{{\cos }^{2}}\dfrac{\theta }{2}$ in the numerator of the L.H.S and $\sin \theta =2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$in the denominator of the L.H.S.
On substituting $1+\cos \theta =2{{\cos }^{2}}\dfrac{\theta }{2}$ and $\sin \theta =2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$ in the left hand side of the equation, we get
L.H.S $={{\cot }^{-1}}\left( \dfrac{2{{\cos }^{2}}\dfrac{x}{2}}{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}} \right)$
$={{\cot }^{-1}}\left( \cot \dfrac{x}{2} \right)$
$=\dfrac{x}{2}$
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 $\dfrac{x}{2}$.
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 $\sqrt{a}+\sqrt{b}$ is $\sqrt{a}-\sqrt{b}$ and not $\left( -\sqrt{a}-\sqrt{b} \right)$. Students generally make this mistake and get their answer wrong.