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.