How do you find the exact value for #tan (sin^-1(-4/5))#?

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Answer 1 · 2016-03-10T10:04:04

#tan(sin^(-1)(-4/5))=+-4/3#

Let #sintheta=(-4/5)# ............(A)
where #theta=sin^(-1)(-4/5)#

Hence as #cos^2theta=1-sin^2theta#, we have

#costheta=sqrt(1-sin^2theta)=sqrt(1-(-4/5)^2# or

#costheta=sqrt(1-16/25)=sqrt(9/25)=+-3/5# ............(B)

Dividing (A) by(B), we get

#tantheta=+-4/3# or

#tan(sin^(-1)(-4/5))=+-4/3#