Question

Prove that:
i) $$\begin{array}{c}{\cos }^{-1}\sqrt{{\displaystyle \frac{2}{3}}}-{\cos }^{-1}{\displaystyle \frac{\sqrt{6}+1}{2\sqrt{3}}}\end{array}$$ = ${\displaystyle \frac{\pi }{6}}$

Answer 1

Hint: Arc of any trigonometry functions represents its inverse function. The inverse function of a number gives the value of angle in radian then it can be changed into degree by multiplying the value of radian to ${\displaystyle \frac{180}{\pi }}$ . the formula for ${\cos }^{-1}x+{\cos }^{-1}y$ is $\left(xy+\sqrt{1-x^2}\sqrt{1-y^2}\right)$ .

Complete step-by-step answer:
Given equation is,
 $$\begin{array}{c}{\cos }^{-1}\sqrt{{\displaystyle \frac{2}{3}}}-{\cos }^{-1}{\displaystyle \frac{\sqrt{6}+1}{2\sqrt{3}}}\end{array}$$
Evaluate further
 $$\begin{array}{l}={\cos }^{-1}\sqrt{{\displaystyle \frac{2}{3}}}-{\cos }^{-1}{\displaystyle \frac{\sqrt{6}+1}{2\sqrt{3}}}\\ ={\cos }^{-1}\left(\sqrt{{\displaystyle \frac{2}{3}}}\times {\displaystyle \frac{\sqrt{6}+1}{2\sqrt{3}}}+\sqrt{1-{\left(\sqrt{{\displaystyle \frac{2}{3}}}\right)}^2}\sqrt{1-{\left({\displaystyle \frac{\sqrt{6}+1}{2\sqrt{3}}}\right)}^2}\right)\\ ={\cos }^{-1}\left(\sqrt{{\displaystyle \frac{2}{3}}}\times {\displaystyle \frac{\sqrt{6}+1}{2\sqrt{3}}}+\sqrt{1-{\displaystyle \frac{2}{3}}}\sqrt{1-\left({\displaystyle \frac{7+2\sqrt{6}}{12}}\right)}\right)\end{array}$$
 $$\begin{array}{l}={\cos }^{-1}\left({\displaystyle \frac{\sqrt{6}+1}{3\sqrt{2}}}+\sqrt{{\displaystyle \frac{1}{3}}}\sqrt{\left({\displaystyle \frac{12-7-2\sqrt{6}}{12}}\right)}\right)\\ ={\cos }^{-1}\left({\displaystyle \frac{\sqrt{6}+1}{3\sqrt{2}}}+\sqrt{{\displaystyle \frac{1}{3}}}\sqrt{\left({\displaystyle \frac{5-2\sqrt{6}}{12}}\right)}\right)\end{array}$$
Here, we can write $$5-2\sqrt{6}$$ in the form of ${\left(a-b\right)}^2$
 $$\begin{array}{l}={\cos }^{-1}\left({\displaystyle \frac{\sqrt{6}+1}{3\sqrt{2}}}+\sqrt{{\displaystyle \frac{1}{3}}}\sqrt{\left({\displaystyle \frac{{\left(\sqrt{3}-\sqrt{2}\right)}^2}{12}}\right)}\right)\\ ={\cos }^{-1}\left({\displaystyle \frac{\sqrt{6}+1}{3\sqrt{2}}}+\sqrt{{\displaystyle \frac{1}{3}}}\times {\displaystyle \frac{\left(\sqrt{3}-\sqrt{2}\right)}{2\sqrt{3}}}\right)\\ ={\cos }^{-1}\left({\displaystyle \frac{\sqrt{6}+1}{3\sqrt{2}}}+{\displaystyle \frac{\left(\sqrt{3}-\sqrt{2}\right)}{6}}\right)\end{array}$$
We will take LCM of the above expression
 $$\begin{array}{l}\\ ={\cos }^{-1}\left({\displaystyle \frac{\sqrt{6}\sqrt{2}+\sqrt{2}+\sqrt{3}-\sqrt{2}}{6}}\right)\end{array}$$
 $$\begin{array}{l}={\cos }^{-1}\left({\displaystyle \frac{2\sqrt{3}+\sqrt{3}}{6}}\right)\\ ={\cos }^{-1}\left({\displaystyle \frac{3\sqrt{3}}{6}}\right)\\ ={\cos }^{-1}\left({\displaystyle \frac{\sqrt{3}}{2}}\right)\end{array}$$
 $$={\displaystyle \frac{\pi }{6}}$$
Now LHS is equal to the RHS, hence proved

Note: In such types of questions, we take left hand side expression and solve the expression to get the right hand side expression. Apply formula and simplify the expression in order to get the right hand side expression. Use direct values of the inverse function to get the answer in the desired unit.