Question

tan(sin^-1(5/6)-cos^-1(1/5)

Answer 1

can someone help me please????

Answer 2

I assume you mean on the sin and cosine ^-1 actually you mean the invsin or invcos, or as Many of us, we call those

ARCSIN
ARCCCOS

meaning the angle whose sin/cosine is ..

Tan(arcsin (5/6)-ARCcos(1/5))

The easy way is to do this on a calculator.

So, the hard way.

Tan(A-B)= (tanA-TanB)/(1+tanA*tanB) check that formula. Now substituter a and b.

On each angle, draw it, you can figure the tan exactly. For instance on arcsin5/6, you know the opposite side is 5, the hypotensuse is 5, so the adjacent6 side is sqrt (36-25)=sqrt11 so Tan of that angle is 5/sqrt11

Answer 3

I answered the same type of question here
http://www.jiskha.com/display.cgi?id=1334539340

Just change the numbers