Question

Re-write the expression as an algebraic expression in x and y (that is, with no trigonometric functions):

Answer 1

\[\tan (\sin^{-1} x+\cos^{-1} x)\]

Answer 2

When I calculate that I get an undefined expression. Since your question mentions some variable y, I have to ask if you meant one of those x's to be a y.

Answer 3

Sorry, you're right. It should be a y instead of an x in cos^-1.

Answer 4

\[ \tan(\sin^{-1}x + \cos^{-1}y) = \frac{\sin(\sin^{-1}x + \cos^{-1} y)}{\cos(\sin^{-1}x + \cos^{-1}y)} \]
\[\sin(\sin^{-1}x + \cos^{-1}y) = \sin(\sin^{-1}x)\cos(\cos^{-1}y) + \sin(\cos^{-1}y)\cos(\sin^{-1}x)\]
\[ = xy + \sqrt{1-\cos^2(\cos^{-1}y)}\sqrt{1-\sin^2(\sin^{-1}x)} = xy + \sqrt{1-y^2}\sqrt{1-x^2}\]
simiarly, the cosines can be rewritten to yield
\[ \cos(\sin^{-1}x + \cos^{-1}y) = \sqrt{1-x^2}\cdot y - x\cdot \sqrt{1-y^2}\]put together, that yields
\[ \frac{xy + \sqrt{1-x^2}\sqrt{1-y^2}}{y\sqrt{1-x^2} - x\sqrt{1-y^2}} \]