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3.Trigonometrical Ratios, Functions and Identities
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$\tan {3^o} + 2\tan {6^o} + 4\tan {12^o} + 8\cot {24^o} = \cot {\theta ^o}$ then
A
$cot (10\ \theta )^o =1$
B
$cot (15\ \theta )^o =1$
C
$\cot {\theta ^o} = 0$
D
$\cot {\left( {15\ \theta } \right)^o} = \sqrt 3$
Solution
$\tan 3^{\circ}+2 \tan 6^{\circ}+4 \tan 12^{\circ}+\frac{8}{\tan 24^{\circ}}$
$\tan 3^{\circ}+2 \tan 6^{\circ}+4 \tan 12^{\circ}+\frac{8\left(1-\tan ^{2} 12^{\circ}\right)}{2 \tan 12^{\circ}}$
$\tan 3^{\circ}+2 \tan 6^{\circ}+\frac{4}{\tan 12^{\circ}}$
$=\tan 3^{\circ}+2 \tan 6^{\circ}+\frac{2\left(1-\tan ^{2} 6^{\circ}\right)}{\tan 6^{\circ}}$
$=\cot 3 \Rightarrow \theta=3^{\circ}$
Standard 11
Mathematics
