Given pi/2 <a<pi, sin a=3/5, what is tan a?
The problem provides the information that angle a is in quadrant 2, where values of tangent function are negative.
You may use the following trigonometric identities,such that:
`cos a = +-sqrt(1 - sin^2 a)`
`tan a = (sin a)/(cos a)`
Reasoning by analogy, yields:
`cos a = +-sqrt(1 - sin^2 a) => cos a = +-sqrt(1 - (3/5)^2)`
Since the values of cosine function are negative in quadrant 2, yields:
`cos a = -sqrt(1 - 9/25) => cos a = -sqrt(16/25)`
`cos a = -4/5`
You need to replace `3/5` for `sin a` and -`4/5` for `cos a` , such that:
`tan a = (3/5)/(-4/5) => tan a = -3/4`
Hence, evaluating the tangent tan a, under the given conditions, yields `tan a = -3/4.`