Given pi/2 <a<pi, sin a=3/5, what is tan a?



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Posted on (Answer #1)

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.`


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