Question.
Choose the correct option and justify your choice:
(i) $\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}=$
(A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30°
(ii) $\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}=$
(A) tan 90° (B) 1 (C) sin 45° (D) 0
(iii) $\sin 2 \mathrm{~A}=2 \sin \mathrm{A}$ is true when $\mathrm{A}=$
(A) 0° (B) 30° (C) 45° (D) 60°
(iv) $\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}=$
(A) cos 60° (B) sin 60 (C) tan 60° (D) sin 30°
Choose the correct option and justify your choice:
(i) $\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}=$
(A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30°
(ii) $\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}=$
(A) tan 90° (B) 1 (C) sin 45° (D) 0
(iii) $\sin 2 \mathrm{~A}=2 \sin \mathrm{A}$ is true when $\mathrm{A}=$
(A) 0° (B) 30° (C) 45° (D) 60°
(iv) $\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}=$
(A) cos 60° (B) sin 60 (C) tan 60° (D) sin 30°
Solution:
(i) Option (A) is correct
$\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}=\frac{2\left(\frac{1}{\sqrt{3}}\right)}{1+\left(\frac{1}{\sqrt{3}}\right)^{2}}=\frac{\frac{2}{\sqrt{3}}}{1+\frac{1}{3}}$
$=\frac{2}{\sqrt{3}} \times \frac{3}{4}=\frac{\sqrt{3}}{2}=\sin 60^{\circ}$
(ii) Option (D) is correct.
$\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}=\frac{1-1}{1+1}=0$
(iii) Option (A) is correct.
sin 2A = 2sinA
2sinA . cosA = 2sinA
cosA = 1
A = 0°
(iv) Option (C) is correct
$\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}$
$=\frac{2 \times \frac{1}{\sqrt{3}}}{1-\left(\frac{1}{\sqrt{3}}\right)^{2}}=\frac{\frac{2}{\sqrt{3}}}{1-\frac{1}{3}}=\frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}=\frac{3}{\sqrt{3}}=\sqrt{3}$
$=\tan 60^{\circ}$
(i) Option (A) is correct
$\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}=\frac{2\left(\frac{1}{\sqrt{3}}\right)}{1+\left(\frac{1}{\sqrt{3}}\right)^{2}}=\frac{\frac{2}{\sqrt{3}}}{1+\frac{1}{3}}$
$=\frac{2}{\sqrt{3}} \times \frac{3}{4}=\frac{\sqrt{3}}{2}=\sin 60^{\circ}$
(ii) Option (D) is correct.
$\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}=\frac{1-1}{1+1}=0$
(iii) Option (A) is correct.
sin 2A = 2sinA
2sinA . cosA = 2sinA
cosA = 1
A = 0°
(iv) Option (C) is correct
$\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}$
$=\frac{2 \times \frac{1}{\sqrt{3}}}{1-\left(\frac{1}{\sqrt{3}}\right)^{2}}=\frac{\frac{2}{\sqrt{3}}}{1-\frac{1}{3}}=\frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}=\frac{3}{\sqrt{3}}=\sqrt{3}$
$=\tan 60^{\circ}$