Question

The value of \(\text{cosec}10^\circ - \sqrt{3} \sec10^\circ\) is equal to:

• A. 8
• B. 2
• C. 6
• D. 4

Hint:

Expressions of the form \(a\sin\theta + b\cos\theta\) are very common in trigonometry problems.
Always try to convert them to the form \(R\sin(\theta \pm \alpha)\) or \(R\cos(\theta \pm \alpha)\) by factoring out \(R = \sqrt{a^2+b^2}\).
This trick simplifies the expression into a single trigonometric function.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to find the numerical value of the trigonometric expression \(\text{cosec}10^\circ - \sqrt{3} \sec10^\circ\).
Step 2: Key Formula or Approach:
The approach is to convert the expression into terms of sine and cosine and then simplify it using standard trigonometric identities, particularly the sum/difference and double angle formulas.
1. Convert cosec and sec to sin and cos: \(\text{cosec}\theta = \frac{1}{\sin\theta}\), \(\sec\theta = \frac{1}{\cos\theta}\).
2. Combine the terms into a single fraction.
3. Use the identity \(A\cos\theta + B\sin\theta = R\cos(\theta \mp \alpha)\).
4. Use the sine double angle formula: \(\sin(2\theta) = 2\sin\theta\cos\theta\).
Step 3: Detailed Explanation:
Let the given expression be \(E\). \[ E = \text{cosec}10^\circ - \sqrt{3} \sec10^\circ \] Convert to sine and cosine: \[ E = \frac{1}{\sin10^\circ} - \frac{\sqrt{3}}{\cos10^\circ} \] Combine the terms by taking a common denominator: \[ E = \frac{\cos10^\circ - \sqrt{3}\sin10^\circ}{\sin10^\circ\cos10^\circ} \] Now, let's simplify the numerator. We can write it in the form \(R\cos(\theta+\alpha)\). Here, \(\cos10^\circ - \sqrt{3}\sin10^\circ\). We can factor out 2: \[ \text{Numerator} = 2 \left( \frac{1}{2}\cos10^\circ - \frac{\sqrt{3}}{2}\sin10^\circ \right) \] We know that \(\cos60^\circ = \frac{1}{2}\) and \(\sin60^\circ = \frac{\sqrt{3}}{2}\). \[ \text{Numerator} = 2 (\cos60^\circ\cos10^\circ - \sin60^\circ\sin10^\circ) \] Using the identity \(\cos(A+B) = \cos A \cos B - \sin A \sin B\): \[ \text{Numerator} = 2 \cos(60^\circ+10^\circ) = 2\cos70^\circ \] Now let's simplify the denominator using the sine double angle formula: \(\sin(2\theta) = 2\sin\theta\cos\theta\). \[ \text{Denominator} = \sin10^\circ\cos10^\circ = \frac{1}{2}(2\sin10^\circ\cos10^\circ) = \frac{1}{2}\sin(2 \times 10^\circ) = \frac{1}{2}\sin20^\circ \] Now, substitute the simplified numerator and denominator back into the expression for E: \[ E = \frac{2\cos70^\circ}{\frac{1}{2}\sin20^\circ} = \frac{4\cos70^\circ}{\sin20^\circ} \] Using the complementary angle identity, \(\cos\theta = \sin(90^\circ-\theta)\): \[ \cos70^\circ = \sin(90^\circ-70^\circ) = \sin20^\circ \] \[ E = \frac{4\sin20^\circ}{\sin20^\circ} = 4 \] Step 4: Final Answer:
The value of the expression is 4.