Question

\[\text{If }y={{\cot }^{-1}}\left( \sqrt{\cos x} \right)-{{\tan }^{-1}}\sqrt{\cos x}.\text{ Prove that }\sin y={{\tan }^{2}}\dfrac{x}{2}\]

Answer 1

Hint: Use the trigonometric identities in inverse trigonometry to simplify the equation. You can use all the formulas you know to make the solution into the equation in terms of tan. Now, convert cot into tan as their sum is \[\dfrac{\pi }{2}\]. Then, apply sin to y to get the value of sin y which is the required result.

Complete step-by-step answer:
Given the equation of y in the question is written as:
\[y={{\cot }^{-1}}\left( \sqrt{\cos x} \right)-{{\tan }^{-1}}\left( \sqrt{\cos x} \right)\]
By basic knowledge of inverse trigonometry, we know the relation:
\[{{\cot }^{-1}}x=\dfrac{\pi }{2}-{{\tan }^{-1}}x....\left( i \right)\]
By substituting this here, we get it as given below:
\[y=\dfrac{\pi }{2}-{{\tan }^{-1}}\left( \sqrt{\cos x} \right)-{{\tan }^{-1}}\left( \sqrt{\cos x} \right)\]
By simplifying the above equation of y, we get,
\[y=\dfrac{\pi }{2}-2{{\tan }^{-1}}\sqrt{\cos x}\]
By basic knowledge of inverse trigonometry, we know the relation:
\[2{{\tan }^{-1}}x={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)\]
By substituting this into our equation, we get it as,
\[y=\dfrac{\pi }{2}-{{\tan }^{-1}}\left( \dfrac{2\sqrt{\cos x}}{1-{{\left( \sqrt{\cos x} \right)}^{2}}} \right)\]
By again using relation (i), we can write it in the form:
\[y={{\cot }^{-1}}\left( \dfrac{2\sqrt{\cos x}}{1-\cos x} \right)\]
By applying cot on both sides, we can write it as:
\[\cot y=\dfrac{2\sqrt{\cos x}}{1-\cos x}\]
By squaring both the sides of the equation, we get,
\[{{\cot }^{2}}y={{\left( \dfrac{2\sqrt{\cos x}}{1-\cos x} \right)}^{2}}\]
By adding 1 on both the sides and simplifying, we get,
\[{{\cot }^{2}}y+1=\dfrac{4\cos x}{{{\left( 1-\cos x \right)}^{2}}}+1=\dfrac{4\cos x+{{\left( 1-\cos x \right)}^{2}}}{{{\left( 1-\cos x \right)}^{2}}}\]
By basic trigonometric knowledge, we know the relation of cot as
\[{{\operatorname{cosec}}^{2}}x={{\cot }^{2}}x+1\]
By algebra, we can say,
\[{{\left( a-b \right)}^{2}}+4ab={{\left( a+b \right)}^{2}}\]
By substituting these values, we get,
\[{{\operatorname{cosec}}^{2}}y=\dfrac{{{\left( 1+\cos x \right)}^{2}}}{{{\left( 1-\cos x \right)}^{2}}}\]
As, \[{{\operatorname{cosec}}^{2}}x=\dfrac{1}{{{\sin }^{2}}x}\], we can write the equation as
\[{{\sin }^{2}}y={{\left( \dfrac{1-\cos x}{1+\cos x} \right)}^{2}}\]
By basic trigonometry, we have the equation as,
\[1-\cos x=2{{\sin }^{2}}\dfrac{x}{2};1+\cos x=2{{\cos }^{2}}\dfrac{x}{2}\]
By substituting these values and applying square root, we get it as,
\[\sin y=\dfrac{2{{\sin }^{2}}\dfrac{x}{2}}{2{{\cos }^{2}}\dfrac{x}{2}}\]
By simplifying the above equation, we can write sin y as,
\[\sin y={{\tan }^{2}}\dfrac{x}{2}\]
Hence proved.

Note: Alternately, at first you can convert tan into cot and apply \[2{{\cot }^{-1}}\] formula then also you will get the same result. You can also substitute \[{{\cos }^{2}}\dfrac{x}{2},{{\sin }^{2}}\dfrac{x}{2}\text{ at }{{\operatorname{cosec}}^{2}}y\] the step itself. We did that +1 step for 2 reasons – (i) to get \[{{\operatorname{cosec}}^{2}}y\text{ from }{{\cot }^{2}}y\], (ii) to get 1 + cos x term from 4 cos x. So, these points are very important. In trigonometric proofs, your manipulation should be in a way that it takes you to the desired result.