KCET · Maths · Trigonometric Ratios & Identities
\( \sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}= \)
- A \( 03 \)
- B \( 11 \)
- C \( 12 \)
- D \( 04 \)
Answer & Solution
Correct Answer
(D) \( 04 \)
Step-by-step Solution
Detailed explanation
(D)
\( \sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ} \)
\( =\frac{\sqrt{3}}{\sin 20^{\circ}}-\frac{1}{\cos 20^{\circ}}=\frac{\sqrt{3} \cos 20^{\circ}-\sin 20^{\circ}}{\sin 20^{\circ} \cdot \cos 20^{\circ}}=\frac{2\left[\frac{\sqrt{3}}{2} \cos 20^{\circ}-\frac{1}{2} \sin 20^{\circ}\right]}{\frac{1}{2}\left[2 \sin 20^{\circ} \cdot \cos 20^{\circ}\right]} \) \( =4 \frac{\left[\sin 60^{\circ} \cdot \cos 20^{\circ}-\cos 60^{\circ} \cdot \sin 20^{\circ}\right]}{\sin 40^{\circ}}=4 \frac{\sin \left(60^{\circ}-20^{\circ}\right)}{\sin 40^{\circ}}=4 \frac{\sin 40^{\circ}}{\sin 40^{\circ}}=4 \)
\( \sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ} \)
\( =\frac{\sqrt{3}}{\sin 20^{\circ}}-\frac{1}{\cos 20^{\circ}}=\frac{\sqrt{3} \cos 20^{\circ}-\sin 20^{\circ}}{\sin 20^{\circ} \cdot \cos 20^{\circ}}=\frac{2\left[\frac{\sqrt{3}}{2} \cos 20^{\circ}-\frac{1}{2} \sin 20^{\circ}\right]}{\frac{1}{2}\left[2 \sin 20^{\circ} \cdot \cos 20^{\circ}\right]} \) \( =4 \frac{\left[\sin 60^{\circ} \cdot \cos 20^{\circ}-\cos 60^{\circ} \cdot \sin 20^{\circ}\right]}{\sin 40^{\circ}}=4 \frac{\sin \left(60^{\circ}-20^{\circ}\right)}{\sin 40^{\circ}}=4 \frac{\sin 40^{\circ}}{\sin 40^{\circ}}=4 \)
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