Q268 / 270 · NDA 2013 (Phase 2) · Mathematics

What is $\frac{\cos 54^{\circ}}{\tan 36^{\circ}}+\frac{\tan 20^{\circ}}{\cot 70^{\circ}}$ equal to?

  • A
    0
  • B
    1
  • Correct answer
    C
    2
  • D
    3

Solution

$\begin{aligned} & \frac{\cot 54^{\circ}}{\tan 36^{\circ}}+\frac{\tan 20^{\circ}}{\cot 70^{\circ}} \\ & \frac{\tan \left(90^{\circ}-54^{\circ}\right)}{\tan 36^{\circ}}+\frac{\tan 20^{\circ}}{\tan \left(90^{\circ}-70^{\circ}\right)} \\ & \frac{\tan 36^{\circ}}{\tan 36^{\circ}}+\frac{\tan 20^{\circ}}{\tan 20^{\circ}}=1+1=2\end{aligned}$

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

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