Question

Find the value of \[ \tan\!\left[\left(2\sin^{-1}\frac{2}{\sqrt{13}}\right)-2\cos^{-1}\!\left(\frac{3}{\sqrt{10}}\right)\right] \]

• A. \(\dfrac{31}{56}\)
• B. \(\dfrac{29}{56}\)
• C. \(\dfrac{33}{56}\)
• D. \(\dfrac{37}{56}\)

Hint:

When inverse trigonometric functions appear inside \(\tan(\cdot)\), first convert them into simple angles and use \(\tan(2\theta)\) identities before applying \(\tan(A-B)\).

Answer 1

The Correct Option is C

Step 1: Evaluate
\(\displaystyle 2\sin^{-1}\frac{2}{\sqrt{13}}\) Let \(\sin\alpha=\dfrac{2}{\sqrt{13}}\). Then \(\cos\alpha=\dfrac{3}{\sqrt{13}}\). \[ \tan\alpha=\frac{2}{3} \] Using the identity: \[ \tan(2\alpha)=\frac{2\tan\alpha}{1-\tan^2\alpha} =\frac{2\cdot\frac{2}{3}}{1-\left(\frac{2}{3}\right)^2} =\frac{\frac{4}{3}}{\frac{5}{9}} =\frac{12}{5} \] Hence, \[ \tan\!\left(2\sin^{-1}\frac{2}{\sqrt{13}}\right)=\frac{12}{5} \]
Step 2: Evaluate
\(\displaystyle 2\cos^{-1}\!\left(\frac{3}{\sqrt{10}}\right)\) Let \(\cos\beta=\dfrac{3}{\sqrt{10}}\). Then \(\sin\beta=\dfrac{1}{\sqrt{10}}\). \[ \tan\beta=\frac{1}{3} \] Using: \[ \tan(2\beta)=\frac{2\tan\beta}{1-\tan^2\beta} =\frac{2\cdot\frac{1}{3}}{1-\left(\frac{1}{3}\right)^2} =\frac{\frac{2}{3}}{\frac{8}{9}} =\frac{3}{4} \] Thus, \[ \tan\!\left(2\cos^{-1}\frac{3}{\sqrt{10}}\right)=\frac{3}{4} \]
Step 3: Use the identity for difference
Let: \[ A=2\sin^{-1}\frac{2}{\sqrt{13}}, \quad B=2\cos^{-1}\frac{3}{\sqrt{10}} \] \[ \tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\,\tan B} \] \[ =\frac{\frac{12}{5}-\frac{3}{4}}{1+\frac{12}{5}\cdot\frac{3}{4}} =\frac{\frac{48-15}{20}}{1+\frac{36}{20}} =\frac{\frac{33}{20}}{\frac{56}{20}} =\frac{33}{56} \] \[ \boxed{\dfrac{33}{56}} \]