Considering the principal values of inverse trigonometric functions, the value of the expr…

Question

Considering the principal values of inverse trigonometric functions, the value of the expression$$\tan \left( 2 \sin^{-1}\left( \frac{2}{\sqrt{13}} \right) - 2 \cos^{-1}\left( \frac{3}{\sqrt{10}} \right) \right)$$is equal to :

Accepted Answer

Simplify the individual components

Let $\alpha ={\sin }^{-1}(\frac{2}{\sqrt{13}})$ and $\beta ={\cos }^{-1}(\frac{3}{\sqrt{10}})$.

For $\alpha $ : Since $\sin \alpha =\frac{2}{\sqrt{13}}$, using Pythagoras' theorem (base $=\sqrt{(\sqrt{13}{)}^{2}-{2}^{2}}=3$ ), we get:

$	an \alpha =\frac{2}{3}$

For $\beta $ : Since $\cos \beta =\frac{3}{\sqrt{10}}$, using Pythagoras' theorem (perpendicular $=\sqrt{(\sqrt{10}{)}^{2}-{3}^{2}}=1$ ), we get:

$	an \beta =\frac{1}{3}$

Rewrite the expression

The expression becomes: $	an (2\alpha -2\beta )=	an [2(\alpha -\beta )]$.

First, find $	an (\alpha -\beta )$ using the formula $	an (A-B)=\frac{	an A-	an B}{1+	an A	an B}$

$	an (\alpha -\beta )=\frac{\frac{2}{3}-\frac{1}{3}}{1+(\frac{2}{3}\cdot \frac{1}{3})}=\frac{\frac{1}{3}}{1+\frac{2}{9}}=\frac{\frac{1}{3}}{\frac{11}{9}}=\frac{1}{3}\cdot \frac{9}{11}=\frac{3}{11}$

Calculate the final value

Now, use the double angle formula $	an (2	heta )=\frac{2	an 	heta }{1-{	an }^{2}	heta }$, where $	heta =(\alpha -\beta )$ :

$\begin{gathered} 	an [2(\alpha -\beta )]=\frac{2(\frac{3}{11})}{1-{(\frac{3}{11})}^{2}} \ 	an [2(\alpha -\beta )]=\frac{\frac{6}{11}}{1-\frac{9}{121}}=\frac{\frac{6}{11}}{\frac{112}{121}} \ 	an [2(\alpha -\beta )]=\frac{6}{11}\cdot \frac{121}{112}=\frac{6\cdot 11}{112}=\frac{66}{112}=\frac{33}{56} \end{gathered}$

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