Solveeit Logo
Login

Question

Mathematics Question on Trigonometric Functions

The value of 3sin⁡15∘−1cos⁡15∘\frac{\sqrt{3}}{\sin15^{\circ}} - \frac{\sqrt{1}}{\cos15^{\circ}} is equal to

A

424\sqrt{ 2}

B

222\sqrt{ 2}

C

2\sqrt{ 2}

D

12\frac{1 }{ \sqrt{ 2}}

Answer

424\sqrt{ 2}

Explanation

Solution

3sin⁡15∘−1cos⁡15∘=3cos⁡15∘−sin⁡15∘sin⁡15∘cos⁡15∘\frac{\sqrt{3}}{\sin 15^{\circ}}-\frac{1}{\cos 15^{\circ}}=\frac{\sqrt{3} \cos 15^{\circ}-\sin 15^{\circ}}{\sin 15^{\circ} \cos 15^{\circ}}
=32cos⁡15∘−12sin⁡15∘12sin⁡15∘cos⁡15∘=\frac{\frac{\sqrt{3}}{2} \cos 15^{\circ}-\frac{1}{2} \sin 15^{\circ}}{\frac{1}{2} \sin 15^{\circ} \cos 15^{\circ}}
=sin⁡60∘cos⁡15∘−cos⁡60∘sin⁡15∘14(sin⁡30∘)=\frac{\sin 60^{\circ} \cos 15^{\circ}-\cos 60^{\circ} \sin 15^{\circ}}{\frac{1}{4}\left(\sin 30^{\circ}\right)}
=sin⁡(60∘−15∘)14×12=sin⁡45∘18=8×12=8×22=42=\frac{\sin \left(60^{\circ}-15^{\circ}\right)}{\frac{1}{4} \times \frac{1}{2}}=\frac{\sin 45^{\circ}}{\frac{1}{8}}=8 \times \frac{1}{\sqrt{2}}=\frac{8 \times \sqrt{2}}{2}=4 \sqrt{2}