Question

If \(2sin^{-1}x-3cos^{-1}x = 4\), then \(2sin^{-1}x+3cos^{-1}x =\)

• A. \(\frac{6π-4}{5}\)
• B. \(\frac{6π+4}{5}\)
• C. \(\frac{5π-4}{6}\)
• D. \(\frac{5π+4}{6}\)

Hint:

Use sin inverse x + cos inverse x = pi/2 to get both unknowns.

Answer 1

The Correct Option is A

Step 1: Approach
Let $P=\sin^{-1}x$ and $Q=\cos^{-1}x$ and treat $P+Q=\dfrac\pi2$ and $2P-3Q=4$ as two linear equations.

Step 2: Solve
From the first, $P=\dfrac\pi2-Q$. Then $\pi-2Q-3Q=4$, so $Q=\dfrac{\pi-4}{5}$ and $P=\dfrac\pi2-\dfrac{\pi-4}{5}=\dfrac{3\pi+8}{10}$.

Step 3: Evaluate
\[ 2P+3Q=\frac{3\pi+8}{5}+\frac{3\pi-12}{5}=\frac{6\pi-4}{5} \]
So the answer is option (A).

Final Answer:
Using the identity for sin inverse plus cos inverse, the value is $\dfrac{6\pi-4}{5}$, option (A). \[ \boxed{\frac{6\pi-4}{5}} \]