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:

To bring -11 into the cot-inverse principal range (0, pi), work out how many multiples of pi need to be added using the floor of (-11/pi) rather than guessing. For the tan-inverse terms, sketching a quick right triangle for the angle often gets you sine and cosine faster than recalling the double-angle-in-terms-of-tan formula.

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: \[ \boxed{\mathrm{(C)\ }4\pi+7} \]

Answer 2

Concept:

• For $\cot^{-1}(\cot x)$, the exact multiple of $\pi$ to remove can be found directly with the floor-function rule $\cot^{-1}(\cot x) = x - \pi\lfloor x/\pi \rfloor$, which always lands the result inside the principal range $(0,\pi)$ without trial and error.
• For $\sin(2\tan^{-1}x)$, drawing a right triangle for the angle $\theta=\tan^{-1}x$ (opposite $=x$, adjacent $=1$, hypotenuse $=\sqrt{1+x^2}$) gives $\sin\theta$ and $\cos\theta$ directly, avoiding the need to recall the double-angle-in-terms-of-tan formula.

Step 1: Reduce $\cot^{-1}(\cot(-11))$ using the floor-function rule
$-11/\pi \approx -3.501$, so $\lfloor -11/\pi \rfloor = -4$ (the next integer below $-3.501$).
$\cot^{-1}(\cot(-11)) = -11 - \pi(-4) = 4\pi - 11$
Check: $4\pi - 11 \approx 12.566 - 11 = 1.566$, and $0 < 1.566 < \pi \approx 3.1416$, so this is indeed the principal value.

Step 2: Evaluate $10\sin\left(2\cos^{-1}(1/\sqrt2)\right)$
$\cos^{-1}(1/\sqrt2) = \pi/4$, a standard angle.
$10\sin(2 \times \pi/4) = 10\sin(\pi/2) = 10 \times 1 = 10$

Step 3: Evaluate $10\sin\left(2\tan^{-1}(2)\right)$ using a right triangle
Let $\theta = \tan^{-1}(2)$. Draw a right triangle with opposite side $2$ and adjacent side $1$, so the hypotenuse is $\sqrt{1^2+2^2}=\sqrt5$.
$\sin\theta = \dfrac{2}{\sqrt5}$, $\cos\theta = \dfrac{1}{\sqrt5}$
$\sin(2\theta) = 2\sin\theta\cos\theta = 2 \times \dfrac{2}{\sqrt5} \times \dfrac{1}{\sqrt5} = \dfrac{4}{5}$
$10 \times \dfrac{4}{5} = 8$

Step 4: Add all three terms
$(4\pi - 11) + 10 + 8 = 4\pi + 7$

Final Answer: $4\pi + 7$