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3.Trigonometrical Ratios, Functions and Identities
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If $tan\ 80^o = a$ and $tan47^o = b$, then $tan37^o$ is equal to -
A
$\frac{{\alpha \, - \,\beta }}{{1\, + \,\alpha \beta }}$
B
$\frac{{\alpha \beta \, + \,1}}{{\alpha \, - \,\beta }}$
C
$\frac{{\alpha \beta \, - \,1}}{{\alpha \, + \,\beta }}$
D
$\frac{{\alpha \, + \,\beta }}{{1\, - \,\alpha \beta }}$
Solution
$\tan 80^{\circ}=\alpha=\cot 10^{\circ}=\frac{1}{\tan 10^{\circ}}$
$\tan 47^{\circ}=\beta$
$\tan 37^{\circ}=\tan \left(47^{\circ}-10^{\circ}\right)=\frac{\tan 47^{\circ}-\tan 10^{\circ}}{1+\tan 47^{\circ} \cdot \tan 10^{\circ}}$
$=\frac{\beta-\frac{1}{\alpha}}{1+\frac{\beta}{\alpha}}=\frac{\alpha \beta-1}{\alpha+\beta}$
Standard 11
Mathematics
