How do you find #tan(sin^-1(p/sqrt(p^2+9)))#?

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Answer 1 · 2016-09-01T04:51:56

#p/3#

Let #a = sin^(-1)(p/sqrt(p^2+9))#

# in Q1 if p >0, and Q4, if p < 0.#, for the principal valur.

In both, cos a is positive.

Then, #sin a = p/sqrt(p^2+9)#

# cos a = sqrt(1-p^2/((p^2+9))#

#=3/sqrt(p^2+9)# and the given expression

#tana = sin a/cos a= p/3#.

If general values are considered, a could be in any of the four

quadrants and tan a = #+-p/3#.
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