Question:
The value of
\[
\sin\left[\cos^{-1}\left(\frac{3}{5}\right)+\tan^{-1}(-2)\right]
\]
is
The value of
\[
\sin\left[\cos^{-1}\left(\frac{3}{5}\right)+\tan^{-1}(-2)\right]
\]
is
Show Hint
For inverse trigonometric functions, first construct a right triangle and determine the required sine and cosine values before applying identities.
Updated On: Jun 16, 2026
- \(\dfrac{2}{5\sqrt5}\)
- \(-\dfrac{2}{5\sqrt5}\)
- \(\dfrac{3}{5\sqrt5}\)
- \(-\dfrac{3}{5\sqrt5}\)
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
Use the identity
\[
\sin(A+B)=\sin A\cos B+\cos A\sin B
\]
Step 1: Let \[ A=\cos^{-1}\left(\frac35\right) \] Then \[ \cos A=\frac35 \] Using a right triangle, \[ \sin A=\frac45 \]
Step 2: Let \[ B=\tan^{-1}(-2) \] Then \[ \tan B=-2 \] Since \(B\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\), \[ \cos B=\frac1{\sqrt5}, \qquad \sin B=-\frac2{\sqrt5} \]
Step 3: Apply the addition formula. \[\begin{aligned} \sin(A+B) &=\sin A\cos B+\cos A\sin B\\ &=\frac45\cdot\frac1{\sqrt5} +\frac35\cdot\left(-\frac2{\sqrt5}\right)\\ &=\frac4{5\sqrt5}-\frac6{5\sqrt5}\\ &=-\frac2{5\sqrt5} \end{aligned}\] \[\begin{aligned} \boxed{-\frac2{5\sqrt5}} \end{aligned}\] Hence, option \(\mathbf{(B)}\) is correct.
Step 1: Let \[ A=\cos^{-1}\left(\frac35\right) \] Then \[ \cos A=\frac35 \] Using a right triangle, \[ \sin A=\frac45 \]
Step 2: Let \[ B=\tan^{-1}(-2) \] Then \[ \tan B=-2 \] Since \(B\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\), \[ \cos B=\frac1{\sqrt5}, \qquad \sin B=-\frac2{\sqrt5} \]
Step 3: Apply the addition formula. \[\begin{aligned} \sin(A+B) &=\sin A\cos B+\cos A\sin B\\ &=\frac45\cdot\frac1{\sqrt5} +\frac35\cdot\left(-\frac2{\sqrt5}\right)\\ &=\frac4{5\sqrt5}-\frac6{5\sqrt5}\\ &=-\frac2{5\sqrt5} \end{aligned}\] \[\begin{aligned} \boxed{-\frac2{5\sqrt5}} \end{aligned}\] Hence, option \(\mathbf{(B)}\) is correct.
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