Question 20

Mathematics Trigonometry Simple Identities Hard

The value of \(\tan 9{^{\circ}}-\tan 27{^{\circ}}-\tan 63{^{\circ}}+\tan 81{^{\circ}}\) is equal to

(A) 5
(B) 3
(C) 4
(D) 6
✓ Correct Answer: Option C Verified Solution

📘 Step-by-Step Solution & Detailed Explanation:

\[ \tan 9^\circ - \tan 27^\circ - \tan 63^\circ + \tan 81^\circ \] \[ = \frac{\sin 9^\circ}{\cos 9^\circ} - \frac{\sin 27^\circ}{\cos 27^\circ} - \left( \frac{\sin 63^\circ}{\cos 63^\circ} - \frac{\sin 81^\circ}{\cos 81^\circ} \right) \] \[ = A - B \] Where: \[ A = \frac{\sin 9^\circ \cos 27^\circ - \cos 9^\circ \sin 27^\circ}{\cos 9^\circ \cos 27^\circ} = \frac{-\sin 18^\circ}{\cos 9^\circ \cos 27^\circ} \] \[ = \frac{-2 \sin 9^\circ}{\cos 27^\circ} \] \[ B = \frac{\sin 63^\circ \cos 81^\circ - \sin 81^\circ \cos 63^\circ}{\cos 63^\circ \cos 81^\circ} \] \[ = \frac{-\sin 18^\circ}{\sin 27^\circ \sin 9^\circ} = \frac{-2 \cos 9^\circ}{\sin 27^\circ} \] Hence, \[ A - B = \frac{2 \cos 9^\circ}{\sin 27^\circ} - \frac{2 \sin 9^\circ}{\cos 27^\circ} \] \[ = \frac{2 \left( \cos 9^\circ \cos 27^\circ - \sin 9^\circ \sin 27^\circ \right)}{\sin 27^\circ \cos 27^\circ} \] \[ = \frac{2 \cos 36^\circ}{\frac{\sin 54^\circ}{2}} = \frac{4 \cos 36^\circ}{\sin 54^\circ} \] \[ = 4 \]

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