Option 2 : tan57° cot37°

CT 1: General Knowledge (History Part 1)

17376

10 Questions
20 Marks
12 Mins

**Concept used:**

tanθ = cot(90° - θ)

tanθ = 1/cotθ

**Calculation:**

\(\dfrac{\tan 57^{\circ}+\cot 37^{\circ}}{\tan 33^{\circ}+ \cot 53^{\circ}}\)

⇒ \(\dfrac{\tan 57^{\circ}+\cot 37^{\circ}}{\tan (90^{\circ} - 57^{\circ})+ \cot(90^{\circ} - 37^{\circ})}\)

⇒ \(\dfrac{\tan 57^{\circ}+\cot 37^{\circ}}{\cot57^{\circ}+ \tan37^{\circ}}\)

⇒ \(\dfrac{\tan 57^{\circ}+\cot 37^{\circ}}{{1\over{\tan57^{\circ}}}+ {1\over\cot37^{\circ}}}\)

⇒ \(\dfrac{\tan 57^{\circ}+\cot 37^{\circ}}{\tan 57^{\circ}+\cot 37^{\circ}} \times tan 57° cot 37°\)

**∴ tan57° cot37°**