Question

Evaluate: `sin 18^@/cos 72^@  + sqrt3 [tan 10° tan 30° tan 40° tan 50° tan 80°]`

Answer 1

sin 18° = sin (90° - 72) = cos 72°

tan 10° = cot 80° tan 50° = cot 40°

`=> sin 18^@/sin 18^@ + sqrt3 [tan 80 cos 30 . tan 40 cot 40 . 1/sqrt3]`

`= 1 + sqrt3 . 1/sqrt3 = 2`