Question

The value of \( \tan(1^\circ) + \tan(89^\circ) \) is equal to:

• A. \( \frac{1}{\sin 1^\circ} \)
• B. \( \frac{2}{\sin 2^\circ} \)
• C. \( \frac{2}{\sin 1^\circ} \)
• D. \( \frac{1}{\sin 2^\circ} \)
• E. \( \frac{\sin 2^\circ}{2} \)

Hint:

The sum \( \tan \theta + \cot \theta \) is always equal to \( \frac{2}{\sin 2\theta} \). This is a very useful shortcut for competitive exams.

Answer 1

The Correct Option is B

Concept: We use the complementary angle identity \( \tan(90^\circ - \theta) = \cot \theta \) and the basic definitions of tan and cot in terms of sin and cos.

Step 1: Convert to a common trigonometric function.
\[ \tan 89^\circ = \tan(90^\circ - 1^\circ) = \cot 1^\circ \] The expression becomes: \[ \tan 1^\circ + \cot 1^\circ \]

Step 2: Write in terms of sine and cosine.
\[ \frac{\sin 1^\circ}{\cos 1^\circ} + \frac{\cos 1^\circ}{\sin 1^\circ} = \frac{\sin^2 1^\circ + \cos^2 1^\circ}{\sin 1^\circ \cos 1^\circ} \]

Step 3: Use the Pythagorean identity and double angle formula.
Since \( \sin^2 \theta + \cos^2 \theta = 1 \): \[ \frac{1}{\sin 1^\circ \cos 1^\circ} \] Multiply numerator and denominator by 2 to use \( \sin 2\theta = 2\sin\theta\cos\theta \): \[ \frac{2}{2\sin 1^\circ \cos 1^\circ} = \frac{2}{\sin 2^\circ} \]