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Question: If \[y=\cot^{-1} \left( \sqrt{\cos x} \right) -\tan^{-1} \left( \sqrt{\cos x} \right) \], then prove...

If y=cot⁡−1(cos⁡x)−tan⁡−1(cos⁡x)y=\cot^{-1} \left( \sqrt{\cos x} \right) -\tan^{-1} \left( \sqrt{\cos x} \right) , then prove that sin⁡y=tan⁡2x2\sin y=\tan^{2} \dfrac{x}{2}.

Explanation

Solution

Hint: In this question it is given that y=cot⁡−1(cos⁡x)−tan⁡−1(cos⁡x)y=\cot^{-1} \left( \sqrt{\cos x} \right) -\tan^{-1} \left( \sqrt{\cos x} \right) ,
We have to prove that sin⁡y=tan⁡2x2\sin y=\tan^{2} \dfrac{x}{2}.
So to find the solution we have to first convert the cot⁡−1θ\cot^{-1} \theta into tan⁡−1θ\tan^{-1} \theta and after that we have to use the formula,
tan⁡−1α−tan⁡−1β=tan⁡−1(α−β1+αβ)\tan^{-1} \alpha -\tan^{-1} \beta =\tan^{-1} \left( \dfrac{\alpha -\beta }{1+\alpha \beta } \right) .........(1)
After using it by simplification we are able to find the solution.

Complete step-by-step solution:
Given,
y=cot⁡−1(cos⁡x)−tan⁡−1(cos⁡x)y=\cot^{-1} \left( \sqrt{\cos x} \right) -\tan^{-1} \left( \sqrt{\cos x} \right)
⇒y=tan⁡−1(1cos⁡x)−tan⁡−1(cos⁡x)\Rightarrow y=\tan^{-1} \left( \dfrac{1}{\sqrt{\cos x} } \right) -\tan^{-1} \left( \sqrt{\cos x} \right) [∵cot⁡−1α=tan⁡−11α\because \cot^{-1} \alpha =\tan^{-1} \dfrac{1}{\alpha }]
Now by using the formula (1) we can write the above equation as,
y=tan⁡−1(1cos⁡x−cos⁡x1+1cos⁡x⋅cos⁡x)y=\tan^{-1} \left( \dfrac{\dfrac{1}{\sqrt{\cos x} } -\sqrt{\cos x} }{1+\dfrac{1}{\sqrt{\cos x} } \cdot \sqrt{\cos x} } \right)
⇒y=tan⁡−1(1−(cos⁡x)2cos⁡x1+1)\Rightarrow y=\tan^{-1} \left( \dfrac{\dfrac{1-\left( \sqrt{\cos x} \right)^{2} }{\sqrt{\cos x} } }{1+1} \right)
⇒y=tan⁡−1(1−cos⁡xcos⁡x2)\Rightarrow y=\tan^{-1} \left( \dfrac{\dfrac{1-\cos x}{\sqrt{\cos x} } }{2} \right)
⇒y=tan⁡−1(1−cos⁡x2cos⁡x)\Rightarrow y=\tan^{-1} \left( \dfrac{1-\cos x}{2\sqrt{\cos x} } \right)
⇒tan⁡y=(1−cos⁡x2cos⁡x)\Rightarrow \tan y=\left( \dfrac{1-\cos x}{2\sqrt{\cos x} } \right)
Now as we know that,cot⁡θ=1tan⁡θ\cot \theta =\dfrac{1}{\tan \theta }, therefore the above equation can be written as,
cot⁡y=1tan⁡y\cot y=\dfrac{1}{\tan y}
⇒cot⁡y=1(1−cos⁡x2cos⁡x)\Rightarrow \cot y=\dfrac{1}{\left( \dfrac{1-\cos x}{2\sqrt{\cos x} } \right) }
⇒cot⁡y=2cos⁡x1−cos⁡x\Rightarrow \cot y=\dfrac{2\sqrt{\cos x} }{1-\cos x}............(1)
Now as we know that, csc⁡2θ=1+cot⁡2θ\csc^{2} \theta =1+\cot^{2} \theta
So by using the above formula we can write,
csc⁡2y=1+cot⁡2y\csc^{2} y=1+\cot^{2} y
⇒csc⁡2y=1+(2cos⁡x1−cos⁡x)2\Rightarrow \csc^{2} y=1+\left( \dfrac{2\sqrt{\cos x} }{1-\cos x} \right)^{2} [from equation (1)]
⇒csc⁡2y=1+(2cos⁡x)2(1−cos⁡x)2\Rightarrow \csc^{2} y=1+\dfrac{\left( 2\sqrt{\cos x} \right)^{2} }{\left( 1-\cos x\right)^{2} }
⇒csc⁡2y=1+22cos⁡x(1−cos⁡x)2\Rightarrow \csc^{2} y=1+\dfrac{2^{2}\cos x}{\left( 1-\cos x\right)^{2} }
⇒csc⁡2y=(1−cos⁡x)2+4cos⁡x(1−cos⁡x)2\Rightarrow \csc^{2} y=\dfrac{\left( 1-\cos x\right)^{2} +4\cos x}{\left( 1-\cos x\right)^{2} }..........(2)
As we know that (a−b)2=a2−2ab+b2\left( a-b\right)^{2} =a^{2}-2ab+b^{2}
By using this identity where a=1, b=cos⁡x\cos x, the above equation can be written as,
csc⁡2y=12−2cos⁡x+cos⁡2x+4cos⁡x(1−cos⁡x)2\csc^{2} y=\dfrac{1^{2}-2\cos x+\cos^{2} x+4\cos x}{\left( 1-\cos x\right)^{2} }
⇒csc⁡2y=12−2cos⁡x+4cos⁡x+cos⁡2x(1−cos⁡x)2\Rightarrow \csc^{2} y=\dfrac{1^{2}-2\cos x+4\cos x+\cos^{2} x}{\left( 1-\cos x\right)^{2} }
⇒csc⁡2y=12+2cos⁡x+cos⁡2x(1−cos⁡x)2\Rightarrow \csc^{2} y=\dfrac{1^{2}+2\cos x+\cos^{2} x}{\left( 1-\cos x\right)^{2} }
Now again as we know that a2+2ab+b2=(a+b)2a^{2}+2ab+b^{2}=\left( a+b\right)^{2}
So by using this identity where a=1, b=cos⁡x\cos x, the above equation can be written as,
csc⁡2y=(1+cos⁡x)2(1−cos⁡x)2\csc^{2} y=\dfrac{\left( 1+\cos x\right)^{2} }{\left( 1-\cos x\right)^{2} }
⇒csc⁡2y=(1+cos⁡x1−cos⁡x)2\Rightarrow \csc^{2} y=\left( \dfrac{1+\cos x}{1-\cos x} \right)^{2}
⇒csc⁡y=(1+cos⁡x1−cos⁡x)\Rightarrow \csc y=\left( \dfrac{1+\cos x}{1-\cos x} \right) [Omitting square from the both side]
⇒1sin⁡x=(1+cos⁡x1−cos⁡x)\Rightarrow \dfrac{1}{\sin x} =\left( \dfrac{1+\cos x}{1-\cos x} \right) [∵csc⁡θ=1sin⁡θ\because \csc \theta =\dfrac{1}{\sin \theta }]
⇒sin⁡x=(1−cos⁡x1+cos⁡x)\Rightarrow \sin x=\left( \dfrac{1-\cos x}{1+\cos x} \right)
Now we have to use two trigonometric identities, which are,
1+cos⁡θ=2cos⁡2θ21+\cos \theta =2\cos^{2} \dfrac{\theta }{2} and
1−cos⁡θ=2sin⁡2θ21-\cos \theta =2\sin^{2} \dfrac{\theta }{2}
So by using these identity we can write the above equation as,
sin⁡x=(2sin⁡2x22cos⁡2x2)\sin x=\left( \dfrac{2\sin^{2} \dfrac{x}{2} }{2\cos^{2} \dfrac{x}{2} } \right) [where θ=x\theta =x]
⇒sin⁡x=tan⁡2x2\Rightarrow \sin x=\tan^{2} \dfrac{x}{2}
Hence proved.

Note: While solving we have transformed tan⁡y\tan y into cot⁡y\cot y, this is because as we know that our solution is in the form of sin⁡y\sin y and to convert it into sin⁡y\sin y we need csc⁡y\csc y and so if we transform tan⁡y\tan y into cot⁡y\cot y then we can easily transform cot⁡y\cot y into csc⁡y\csc y by the formula csc⁡2y=1+cot⁡2y\csc^{2} y=1+\cot^{2} y.