Start Learning Free

Before You Begin

Question #41158

Difficulty: 8/10 | 🟢 Free

The Question

Let zk=(1+itan⁡π181−itan⁡π18)kz_k = \left( \frac{1 + i \tan \frac{\pi}{18}}{1 - i \tan \frac{\pi}{18}} \right)^k for k∈Nk \in \mathbb{N}. Find the value of the following expression in degrees: cos⁡−1(Re(z13))+sin⁡−1(Im(z10))+tan⁡−1(Im(z16)Re(z16))\cos^{-1}(\text{Re}(z_{13})) + \sin^{-1}(\text{Im}(z_{10})) + \tan^{-1}\left(\frac{\text{Im}(z_{16})}{\text{Re}(z_{16})}\right)

Options

  • A.

    40∘40^\circ

  • B.

    160∘160^\circ

  • C.

    20∘20^\circ

  • D.

    780∘780^\circ

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Principal Value Branch Boundaries|Algebraic Manipulation of i|Radian-Degree Conversion Identity

Why This Question Is Tricky

Common Mistakes

This question forces a collision between complex number algebra and the geometric constraints of inverse trigonometric functions.

The student must first use Euler's formula to reduce the complex ratio into a standard polar form.

The primary trap lies in the blind application of the identity f−1(f(x))=xf^{-1}(f(x)) = x.

Students who ignore the principal value branches of ITF will incorrectly sum the raw arguments (260∘+200∘+320∘=780∘260^\circ + 200^\circ + 320^\circ = 780^\circ) or make sign errors when reducing to the first quadrant (e.g., yielding 160∘160^\circ).

It rigorously tests domain awareness.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterInverse Trigonometric Functions - Class 12
ConceptPrincipal Value Branch Boundaries
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree