Question

Consider the statements given below
I. The value of $\sin 75^{\circ }$ is $\frac{\sqrt{3}+1}{2\sqrt{2}}$.
II. The value of $\tan 15^{\circ }$ is $2+\sqrt{3}$.
III. $\tan 15^{\circ }=\tan \left(45^{\circ }-30^{\circ }\right)=\frac{\tan 45^{\circ }-\tan 30^{\circ }}{1+\tan 45^{\circ }\cdot \tan 30^{\circ }}$
Choose the correct option.

• A. I is correct
• B. II is correct
• C. I and III are correct
• D. II and III are correct

Answer 1

Answer: (C)

(i) $\sin 75^{\circ }=\sin \left(45^{\circ }+30^{\circ }\right)$
$\sin \left(45^{\circ }+30^{\circ }\right)=\sin 45^{\circ }\cos 30^{\circ }+\cos 45^{\circ }\sin 30^{\circ }$
$[\because \sin (A+B)=\sin A\cos B+\cos A\sin B]$
$=\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\times \frac{1}{2}=\frac{\sqrt{3}+1}{2\sqrt{2}}$
(ii) $\tan 15^{\circ }=\tan \left(45^{\circ }-30^{\circ }\right)$
$=\frac{\tan 45^{\circ }-\tan 30^{\circ }}{1+\tan 45^{\circ }\tan 30^{\circ }}\left[\because \tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}\right]$
$=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}=\frac{\sqrt{3}-1}{\sqrt{3}+1}$
$=\frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{(\sqrt{3}-1{)}^2}{3-1}$
$=\frac{3+1-2\sqrt{3}}{2}=\frac{4-2\sqrt{3}}{2}$
$=2-\sqrt{3}$
(iii) Clearly, it is true.
Hence, only (i) and (iii) are true.