NIMCET 2010 — Mathematics PYQ

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Question

NIMCET 2010 — Mathematics PYQ

NIMCET | Mathematics | 2010

The value of $\sqrt{3} \cot 20^\circ - 4 \cos 20^\circ$ is:

Choose the correct answer:

Explanation

Step 1: Convert the expression into sine and cosine.

We know that $\cot \theta = \frac{\cos \theta}{\sin \theta}$ .

\sqrt{3} \frac{\cos 20^\circ}{\sin 20^\circ} - 4 \cos 20^\circ

Step 2: Find a common denominator.

\frac{\sqrt{3} \cos 20^\circ - 4 \cos 20^\circ \sin 20^\circ}{\sin 20^\circ}

Step 3: Simplify the numerator using the identity $2 \sin \theta \cos \theta = \sin 2\theta$ .

We can rewrite $4 \cos 20^\circ \sin 20^\circ$ as $2(2 \sin 20^\circ \cos 20^\circ) = 2 \sin 40^\circ$ .

\frac{\sqrt{3} \cos 20^\circ - 2 \sin 40^\circ}{\sin 20^\circ}

Step 4: Use the value $\sqrt{3} = 2 \sin 60^\circ$ to facilitate the subtraction.

\frac{2 \sin 60^\circ \cos 20^\circ - 2 \sin 40^\circ}{\sin 20^\circ}

Step 5: Apply the product-to-sum identity $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$ .

Here, $A = 60^\circ$ and $B = 20^\circ$ :

\frac{(\sin(60^\circ + 20^\circ) + \sin(60^\circ - 20^\circ)) - 2 \sin 40^\circ}{\sin 20^\circ}

\frac{\sin 80^\circ + \sin 40^\circ - 2 \sin 40^\circ}{\sin 20^\circ}

\frac{\sin 80^\circ - \sin 40^\circ}{\sin 20^\circ}

Step 6: Apply the difference-to-product identity $\sin C - \sin D = 2 \cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)$ .

\frac{2 \cos\left(\frac{80^\circ+40^\circ}{2}\right) \sin\left(\frac{80^\circ-40^\circ}{2}\right)}{\sin 20^\circ}

\frac{2 \cos 60^\circ \sin 20^\circ}{\sin 20^\circ}

Step 7: Final Calculation.

Since $\cos 60^\circ = \frac{1}{2}$ :

\frac{2 \times \frac{1}{2} \times \sin 20^\circ}{\sin 20^\circ} = \frac{\sin 20^\circ}{\sin 20^\circ} = 1

Correct Option: (b) 1