MCQ
If $tan\ 80^o = a$ and $tan47^o = b$, then $tan37^o$ is equal to -
  • A
    $\frac{{\alpha \, - \,\beta }}{{1\, + \,\alpha \beta }}$
  • B
    $\frac{{\alpha \beta \, + \,1}}{{\alpha \, - \,\beta }}$
  • $\frac{{\alpha \beta \, - \,1}}{{\alpha \, + \,\beta }}$
  • D
    $\frac{{\alpha \, + \,\beta }}{{1\, - \,\alpha \beta }}$

Answer

Correct option: C.
$\frac{{\alpha \beta \, - \,1}}{{\alpha \, + \,\beta }}$
c
$\tan 80^{\circ}=\alpha=\cot 10^{\circ}=\frac{1}{\tan 10^{\circ}}$

$\tan 47^{\circ}=\beta$

$\tan 37^{\circ}=\tan \left(47^{\circ}-10^{\circ}\right)=\frac{\tan 47^{\circ}-\tan 10^{\circ}}{1+\tan 47^{\circ} \cdot \tan 10^{\circ}}$

$=\frac{\beta-\frac{1}{\alpha}}{1+\frac{\beta}{\alpha}}=\frac{\alpha \beta-1}{\alpha+\beta}$

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