Question

$\cos ec15°-sec15°$ is equal to

• A. $2\sqrt{2}$
• B. $\sqrt{6}$
• C. $2\sqrt{6}$
• D. $\sqrt{6}+\sqrt{2}$

Answer 1

The correct option is A

$2\sqrt{2}$

Explanation for the correct option:

Step 1. Given $\cos ec15°–sec15°$

$=\frac{1}{\sin 15°}–\frac{1}{\cos 15°}$

$=\frac{(\cos 15°–\sin 15°)}{(\sin 15°\cos 15°)}$

Step 2. Multiply and divide it by 2,

$=\frac{2(\cos 15°–\sin 15°)}{(2\sin 15°\cos 15°)}$

$=\frac{2(\cos 15°–\sin 15°)}{\sin 30°}$ ; $\left[\mathbf{\because }\mathbf{2}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathbf{2}\mathit{\theta }\right]$

Step 3. Multiply and divide it by $\sqrt{2}$,

$=2\cos ec30°\times \sqrt{2}\times \left[\left(\frac{1}{\sqrt{2}}\right)\cos 15°–\left(\frac{1}{\sqrt{2}}\right)\sin 15°\right]$

$=2\times 2\times \sqrt{2}\times [\sin 45°\cos 15°–\cos 45°\sin 15°]$ ; $\left[\because cosec30°=2,sin45°=\frac{1}{\surd 2},cos45°=-\frac{1}{\surd 2}\right]$

$=4\sqrt{2}\left[\sin \right(45°–15°\left)\right]$ $\left[\mathbf{\because }\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{A}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{B}\mathbf{-}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{A}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{B}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathbf{(}\mathit{A}\mathbf{-}\mathit{B}\mathbf{)}\right]$

$=4\sqrt{2}\sin 30°$

$=4\sqrt{2}\times \frac{1}{2}$

$=\mathbf{2}\sqrt{\mathbf{2}}$

Hence, Option ‘A’ is Correct.