Question

In each of the following, one trigonometric ratio is given. Find the values of the other trigonometric.

sinB = `sqrt(3)/(2)`

Answer 1

sinB = `sqrt(3)/(2)`

sinB = `"Perpendicular"/"Hypotenuse" = sqrt(3)/(2)`

By Pythagoras theorem, we have
(Hypotenuse)² = (Perpendicular)² + (Base)²
⇒ Base = `sqrt(("Hypotenuse")^2 - ("Perpendicular")^2`
⇒ Base
`sqrt((2)^2 - (sqrt(3))^2`
= `sqrt(4 - 3)`
= `sqrt(1)`
= 1

cosB = `"Base"/"Hypotenuse" = (1)/(2)`

tanB = `"Perpendicular"/"Base" = sqrt(3)`

secB = `(1)/"cosB"` = 2

cotB = `(1)/"tanB" = (1)/sqrt(3)`

cosecB= `(1)/"sinA" = (2)/sqrt(3)`.