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Match List I with List II and select the correct answer using the code given below the lists
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Verified Answer
The correct answer is:
A-4 B-2 C-1
(A) $\tan 15^{\circ}=\tan \left(45^{\circ}-30^{\circ}\right)$
$=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}=\frac{\sqrt{3}-1}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{3+1-2 \sqrt{3}}{2}=2-\sqrt{3}$
(B) $\tan 75^{\circ}=\tan \left(45^{\circ}+30^{\circ}\right)=\frac{\tan 45^{\circ}+\tan 30^{\circ}}{1-\tan 45^{\circ} \tan 30^{\circ}}$
(C) $\tan \left(105^{\circ}\right)=\tan \left(60^{\circ}+45^{\circ}\right)$
$=\frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}=\frac{\sqrt{3}+1}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}$ (By Rationaliziang)
$=2+\sqrt{3}$
$=\frac{\sqrt{3}+1}{1-\sqrt{3}} \times \frac{1+\sqrt{3}}{1+\sqrt{3}}=\frac{(\sqrt{3}+1)^{2}}{1-3}$
$=\frac{4+2 \sqrt{3}}{-2}=-2-\sqrt{3}$
Hence, option (b) is correct.
$=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}=\frac{\sqrt{3}-1}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{3+1-2 \sqrt{3}}{2}=2-\sqrt{3}$
(B) $\tan 75^{\circ}=\tan \left(45^{\circ}+30^{\circ}\right)=\frac{\tan 45^{\circ}+\tan 30^{\circ}}{1-\tan 45^{\circ} \tan 30^{\circ}}$
(C) $\tan \left(105^{\circ}\right)=\tan \left(60^{\circ}+45^{\circ}\right)$
$=\frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}=\frac{\sqrt{3}+1}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}$ (By Rationaliziang)
$=2+\sqrt{3}$
$=\frac{\sqrt{3}+1}{1-\sqrt{3}} \times \frac{1+\sqrt{3}}{1+\sqrt{3}}=\frac{(\sqrt{3}+1)^{2}}{1-3}$
$=\frac{4+2 \sqrt{3}}{-2}=-2-\sqrt{3}$
Hence, option (b) is correct.
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