Q93 / 180 · KCET 2012 · Mathematics

$\frac{\sin 70^{\circ}+\cos 40^{\circ}}{\cos 70^{\circ}+\sin 40^{\circ}}$ is equal to

  • A
    $\frac{1}{\sqrt{3}}$
  • Correct answer
    B
    $\sqrt{3}$
  • C
    $\frac{1}{2}$
  • D
    1

Solution

$\begin{aligned} \frac{\sin 70^{\circ}+\cos 40^{\circ}}{\cos 70^{\circ}+\sin 40^{\circ}} \\=& \frac{\sin \left(90^{\circ}-20^{\circ}\right)+\sin \left(90^{\circ}-40^{\circ}\right)}{\cos \left(90^{\circ}-20^{\circ}\right)+\sin \left(90^{\circ}-50^{\circ}\right)} \\=& \frac{\cos 20^{\circ}+\cos 40^{\circ}}{\cos 70^{\circ}+\cos 50^{\circ}} \\=& \frac{2 \cos 30^{\circ} \cos 10^{\circ}}{2 \cos 60^{\circ} \cos 10^{\circ}}=\frac{2 \cos 30^{\circ}}{2 \cos 60^{\circ}} \\=& \frac{2(\sqrt{3} / 2)}{2\left(\frac{1}{2}\right)}=\sqrt{3} \end{aligned}$

Hence, the correct answer is option: 'B'.

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