Question

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

Answer 1

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

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

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