Question

if sin1°=a/b , then tan89°=?

Answer 1

Required answer is in attachment.

Answer 2

Answer:

Given :

Sin1° = a/b

Cos (90°-1°) = a/b

Cos89° = a/b

1/Sec89° = a/b

Sec89° = b/a

Squaring both sides :

Sec^2 89° = b^2/a^2

Using Identity : Sec^2A = 1 + Tan^2A

1 + Tan^2 89° = b^2/a^2

Tan^2 89° = b^2/a^2 -1

Tan89° = √b^2-a^2/a Answer!