Question:
If \(2sin^{-1}x-3cos^{-1}x = 4\), then \(2sin^{-1}x+3cos^{-1}x =\)
If \(2sin^{-1}x-3cos^{-1}x = 4\), then \(2sin^{-1}x+3cos^{-1}x =\)
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Use sin inverse x + cos inverse x = pi/2 to get both unknowns.
Updated On: Oct 1, 2026
- \(\frac{6π-4}{5}\)
- \(\frac{6π+4}{5}\)
- \(\frac{5π-4}{6}\)
- \(\frac{5π+4}{6}\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept
For \(x\in[-1,1]\): \(\sin^{-1}x+\cos^{-1}x=\dfrac\pi2\). Let \(s=\sin^{-1}x\), so \(\cos^{-1}x=\dfrac\pi2-s\).
Step 2: Use the given equation
\[ 2s-3\left(\frac\pi2-s\right)=4\Rightarrow5s=4+\frac{3\pi}{2} \]
\[ s=\frac{8+3\pi}{10} \]
Step 3: Required expression
\[ 2s+3\left(\frac\pi2-s\right)=\frac{3\pi}{2}-s=\frac{3\pi}{2}-\frac{8+3\pi}{10} \]
\[ =\frac{15\pi-8-3\pi}{10}=\frac{12\pi-8}{10}=\frac{6\pi-4}{5} \]
This is option (A).
Final Answer:
Using the identity for sin inverse plus cos inverse, the value is \(\dfrac{6\pi-4}{5}\), option (A). \[ \boxed{\frac{6\pi-4}{5}} \]
For \(x\in[-1,1]\): \(\sin^{-1}x+\cos^{-1}x=\dfrac\pi2\). Let \(s=\sin^{-1}x\), so \(\cos^{-1}x=\dfrac\pi2-s\).
Step 2: Use the given equation
\[ 2s-3\left(\frac\pi2-s\right)=4\Rightarrow5s=4+\frac{3\pi}{2} \]
\[ s=\frac{8+3\pi}{10} \]
Step 3: Required expression
\[ 2s+3\left(\frac\pi2-s\right)=\frac{3\pi}{2}-s=\frac{3\pi}{2}-\frac{8+3\pi}{10} \]
\[ =\frac{15\pi-8-3\pi}{10}=\frac{12\pi-8}{10}=\frac{6\pi-4}{5} \]
This is option (A).
Final Answer:
Using the identity for sin inverse plus cos inverse, the value is \(\dfrac{6\pi-4}{5}\), option (A). \[ \boxed{\frac{6\pi-4}{5}} \]
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