Question

If $ P = \tan 15^\circ + \cot 15^\circ $, $ Q = \tan 22\frac{1}{2}^\circ + \cot 22\frac{1}{2}^\circ $, and $ R = \sin 54^\circ + \sin 18^\circ $, then their ascending order is

• A. P, Q, R
• B. P, R, Q
• C. R, Q, P
• D. R, P, Q

Hint:

Use trigonometric identities to simplify expressions. Remember $ \tan \theta + \cot \theta = 2/\sin(2\theta) $ and sum-to-product formulas.

Answer 1

The Correct Option is C

Step 1: Simplify the expression for $ P $.
$ P = \tan 15^\circ + \cot 15^\circ = \frac{2}{\sin (2 \times 15^\circ)} = \frac{2}{\sin 30^\circ} = \frac{2}{1/2} = 4 $. 
Step 2: Simplify the expression for $ Q $.
$ Q = \tan 22\frac{1}{2}^\circ + \cot 22\frac{1}{2}^\circ = \frac{2}{\sin (2 \times 22\frac{1}{2}^\circ)} = \frac{2}{\sin 45^\circ} = \frac{2}{1/\sqrt{2}} = 2\sqrt{2} \approx 2.828 $. 
Step 3: Simplify the expression for $ R $.
$ R = \sin 54^\circ + \sin 18^\circ = 2 \sin \left(\frac{54^\circ + 18^\circ}{2}\right) \cos \left(\frac{54^\circ - 18^\circ}{2}\right) $ $ R = 2 \sin 36^\circ \cos 18^\circ = 2 \left(\frac{\sqrt{10 - 2\sqrt{5}}}{4}\right) \left(\frac{\sqrt{10 + 2\sqrt{5}}}{4}\right) = \frac{1}{8} \sqrt{100 - 20} = \frac{\sqrt{80}}{8} = \frac{4\sqrt{5}}{8} = \frac{\sqrt{5}}{2} \approx 1.118 $. 
Step 4: Compare the values of $ P $, $ Q $, and $ R $.
$ P = 4 $, $ Q \approx 2.828 $, $ R \approx 1.118 $. 
Step 5: Arrange $ P $, $ Q $, and $ R $ in ascending order.
$ R<Q<P $. 
Step 6: Match the order with the given options.
The ascending order is R, Q, P.