Question

The value of \( \csc 10^\circ - \sqrt{3}\sec 10^\circ \) is:

• A. \(1\)
• B. \(2\)
• C. \(4\)
• D. None of these

Hint:

Use $\csc\theta=\sec\theta\cot\theta$ to factor out $\sec10^\circ$. Then write $\sqrt3$ as $\cot30^\circ$ and apply the cotangent-difference identity.

Answer 1

The Correct Option is C

Step 1: Write the given expression in terms of sine and cosine: \[ \csc 10^\circ - \sqrt{3}\sec 10^\circ = \frac{1}{\sin 10^\circ} - \frac{\sqrt{3}}{\cos 10^\circ} \]
Step 2: Take LCM of the denominators: \[ = \frac{\cos 10^\circ - \sqrt{3}\sin 10^\circ}{\sin 10^\circ \cos 10^\circ} \]
Step 3: Use the identity \[ \cos \theta - \sqrt{3}\sin \theta = 2\cos(\theta + 60^\circ) \] \[ \Rightarrow \cos 10^\circ - \sqrt{3}\sin 10^\circ = 2\cos 70^\circ = 2\sin 20^\circ \]
Step 4: Simplify the denominator: \[ \sin 10^\circ \cos 10^\circ = \frac{1}{2}\sin 20^\circ \]
Step 5: Substitute: \[ = \frac{2\sin 20^\circ}{\frac{1}{2}\sin 20^\circ} = 4 \]

Answer 2

Concept:

• Rewrite the expression as a secant multiplied by a difference of cotangents.
• The identity $\cot A-\cot B=\dfrac{\sin(B-A)}{\sin A\sin B}$ evaluates that difference directly.

Step 1: Factor out $\sec10^\circ$.
Since $\csc10^\circ=\sec10^\circ\cot10^\circ$,
$\csc10^\circ-\sqrt3\sec10^\circ=\sec10^\circ(\cot10^\circ-\sqrt3)$

Step 2: Express $\sqrt3$ as a cotangent.
$\sqrt3=\cot30^\circ$
Thus the bracket becomes $\cot10^\circ-\cot30^\circ$.

Step 3: Use the cotangent-difference identity.
$\cot10^\circ-\cot30^\circ=\dfrac{\sin20^\circ}{\sin10^\circ\sin30^\circ}$
$=\dfrac{2\sin20^\circ}{\sin10^\circ}=\dfrac{2(2\sin10^\circ\cos10^\circ)}{\sin10^\circ}=4\cos10^\circ$

Step 4: Multiply by the factored secant.
$\sec10^\circ(4\cos10^\circ)=4$

Final Answer: $4$