How do you use the properties of inverse trigonometric functions to evaluate $tan (arcsin (0.31))$ $tan(arcsin (0.31))$ ?

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Answer 1 · 2014-11-25T03:16:18

Let

$θ = arcsin (0.31)$ $theta=arcsin(0.31)$

by rewriting in terms of sine,

$sin θ = 0.31 = \frac{O}{H} \Rightarrow$ $sin theta=0.31={O}/H =>$ Let ${(O=0.31),(H=1):}$

By Pythagorean Theorem,

$A=sqrt{H^2-O^2}=sqrt{1^2-(0.31)^2}=sqrt{0.9039}$

Hence,

$tan(arcsin(0.31))=tan theta={O}/A={0.31}/{sqrt{0.9039}}$

I hope that this was helpful.

How do you use the properties of inverse trigonometric functions to evaluate tan(arcsin (0.31))tan(arcsin(0.31))tan(arcsin (0.31))?