Question

Considering only the principal values of the inverse trigonometric functions, evaluate: \[ \cot^{-1}(\cot(-11)) +10\sin\left(2\cos^{-1}\left(\frac1{\sqrt2}\right)\right) +10\sin\left(2\tan^{-1}(2)\right) \]

• A. \(3\pi+7\)
• B. \(7\)
• C. \(4\pi+7\)
• D. \(3\pi-5\)

Hint:

Principal value range: \[ \cot^{-1}x\in(0,\pi) \] Useful identity: \[ \sin(2\theta)=\frac{2\tan\theta}{1+\tan^2\theta} \]

Answer 1

The Correct Option is C

Step 1: Evaluate \(\cot^{-1}(\cot(-11))\).
Principal value range of: \[ \cot^{-1}x \] is: \[ (0,\pi) \] Now: \[ -11+4\pi \] lies in: \[ (0,\pi) \] Thus: \[ \cot^{-1}(\cot(-11)) = -11+4\pi \]

Step 2: Evaluate the second term.
Let: \[ \theta=\cos^{-1}\left(\frac1{\sqrt2}\right) \] Then: \[ \theta=\frac{\pi}{4} \] Hence: \[ 10\sin\left(2\times\frac{\pi}{4}\right) = 10\sin\left(\frac{\pi}{2}\right) \] \[ =10 \]

Step 3: Evaluate the third term.
Using: \[ \sin(2\theta)=\frac{2\tan\theta}{1+\tan^2\theta} \] Let: \[ \theta=\tan^{-1}(2) \] Then: \[ \tan\theta=2 \] Thus: \[ \sin(2\theta) = \frac{2(2)}{1+2^2} = \frac45 \] Therefore: \[ 10\sin(2\theta) = 10\times\frac45 = 8 \]

Step 4: Add all terms.
\[ (-11+4\pi)+10+8 \] \[ =4\pi+7 \]

Step 5: Identify the correct option.
Hence: \[ \boxed{\mathrm{(C)\ }4\pi+7} \]