What is the value of \(\sin^{-1} \frac{4}{5} + 2 \tan^{-1} \frac{1}{3

1.	What is the value of \(\sin^{-1} \frac{4}{5} + 2 \tan^{-1} \frac{1}{3} \ ?\)1. \(\dfrac{\pi}{3}\)2. \(\dfrac{\pi}{2}\)3. \(\dfrac{\pi}{4}\)4. \(\dfrac{\pi}{6}\)
Answer» Correct Answer - Option 2 : \(\dfrac{\pi}{2}\) Concept: \(\rm sin^2x+cos^2x=1\) \(\rm \tan^{-1}(tanx)= x\) \(\rm 2 \tan^{-1} x= tan^{-1}(\frac{2x}{1-x^2})\) Calculation: Let, \(\rm \sin^{-1} \dfrac{4}{5} =x\) sin x= 4/5 Now, \(\rm cos^2x=1-sin^2x=1-\frac{16}{25}=\frac{9}{25}\) ⇒ cos x = 3/5 Now, cot x = cos x/sin x = 3/4 ....(1) \(\rm 2 \tan^{-1} \dfrac{1}{3} =tan^{-1}(\frac{\frac 2 3}{1-\frac 19})\) ....(∵ \(\rm 2 \tan^{-1} x= tan^{-1}(\frac{2x}{1-x^2})\)) \(\rm =tan^{-1}(\frac{\frac 2 3}{\frac 89})\) \(\rm =tan^{-1}(\frac 3 4)\) \(\rm =tan^{-1}(cot x)\) ....(from 1) \(\rm =tan^{-1}(tan (\frac π 2-x))\) = π/2 - x .....(∵ \(\rm \tan^{-1}(tanx)= x\)) = \(\rm \frac \pi 2-\sin^{-1} \dfrac{4}{5} \) ∴ \(\sin^{-1} \dfrac{4}{5} + 2 \tan^{-1} \dfrac{1}{3} = \sin^{-1} \dfrac{4}{5}+\frac\pi 2 -\sin^{-1} \dfrac{4}{5}\) = \(\dfrac{\pi}{2}\) Hence, option (2) is correct.

What is the value of \(\sin^{-1} \frac{4}{5} + 2 \tan^{-1} \frac{1}{3} \ ?\)1. \(\dfrac{\pi}{3}\)2. \(\dfrac{\pi}{2}\)3. \(\dfrac{\pi}{4}\)4. \(\dfrac{\pi}{6}\)

Answer» Correct Answer - Option 2 : \(\dfrac{\pi}{2}\)

Concept:

\(\rm sin^2x+cos^2x=1\)

\(\rm \tan^{-1}(tanx)= x\)

\(\rm 2 \tan^{-1} x= tan^{-1}(\frac{2x}{1-x^2})\)

Calculation:

Let, \(\rm \sin^{-1} \dfrac{4}{5} =x\)

sin x= 4/5

Now, \(\rm cos^2x=1-sin^2x=1-\frac{16}{25}=\frac{9}{25}\)

⇒ cos x = 3/5

Now, cot x = cos x/sin x = 3/4 ....(1)

\(\rm 2 \tan^{-1} \dfrac{1}{3} =tan^{-1}(\frac{\frac 2 3}{1-\frac 19})\) ....(∵ \(\rm 2 \tan^{-1} x= tan^{-1}(\frac{2x}{1-x^2})\))

\(\rm =tan^{-1}(\frac{\frac 2 3}{\frac 89})\)

\(\rm =tan^{-1}(\frac 3 4)\)

\(\rm =tan^{-1}(cot x)\) ....(from 1)

\(\rm =tan^{-1}(tan (\frac π 2-x))\)

= π/2 - x .....(∵ \(\rm \tan^{-1}(tanx)= x\))

= \(\rm \frac \pi 2-\sin^{-1} \dfrac{4}{5} \)

∴ \(\sin^{-1} \dfrac{4}{5} + 2 \tan^{-1} \dfrac{1}{3} = \sin^{-1} \dfrac{4}{5}+\frac\pi 2 -\sin^{-1} \dfrac{4}{5}\)