Step1

Consider the given expression as \(\cos(\sin^{-1}x-cos^{-1}y)\).

\(Let\ \sin^{-1} x = \alpha\ and\ \cos^{-1} y= \beta,\ then\ x=\sin \alpha\ and\ y=\cos \beta\).

Now the given expression can be write as \(\cos(\alpha — \beta)\).

Known formula:

l.\(\cos(\alpha — \beta)=\cos \alpha \cos \beta+\sin \alpha \sin \beta\)

2.\(\sin^{2} 0+ \cos^{2} 0=1\)

Step 2

Compute the value of \(\cos \alpha\ and\ \sin \beta\) as follows.

\(\cos \alpha = \sqrt{1-\sin^{2} \alpha}\)

\(=\sqrt{1-x^{2}}\)

\(\sin \beta = \sqrt{1-\cos^{2} \beta}\)

\(=\sqrt{1-y^{2}}\)

Substitute the values of \(\cos \alpha\ and\ \sin \beta\) in the formula (1).

\(\cos(\alpha — \beta) =\cos \alpha \cos \beta+\sin \alpha \sin \beta\)

\(=(\sqrt{1-x^{2}})(y)+(x)(\sqrt{1-y^{2}})\)

\(=x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}}\)

Thus, the expression \(\cos(\sin^{-1}x-cos^{-1}y)\) as algebraic in x and y as x \(\sqrt{1-y^{2}}+y\sqrt{1-x^{2}}\)

Consider the given expression as \(\cos(\sin^{-1}x-cos^{-1}y)\).

\(Let\ \sin^{-1} x = \alpha\ and\ \cos^{-1} y= \beta,\ then\ x=\sin \alpha\ and\ y=\cos \beta\).

Now the given expression can be write as \(\cos(\alpha — \beta)\).

Known formula:

l.\(\cos(\alpha — \beta)=\cos \alpha \cos \beta+\sin \alpha \sin \beta\)

2.\(\sin^{2} 0+ \cos^{2} 0=1\)

Step 2

Compute the value of \(\cos \alpha\ and\ \sin \beta\) as follows.

\(\cos \alpha = \sqrt{1-\sin^{2} \alpha}\)

\(=\sqrt{1-x^{2}}\)

\(\sin \beta = \sqrt{1-\cos^{2} \beta}\)

\(=\sqrt{1-y^{2}}\)

Substitute the values of \(\cos \alpha\ and\ \sin \beta\) in the formula (1).

\(\cos(\alpha — \beta) =\cos \alpha \cos \beta+\sin \alpha \sin \beta\)

\(=(\sqrt{1-x^{2}})(y)+(x)(\sqrt{1-y^{2}})\)

\(=x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}}\)

Thus, the expression \(\cos(\sin^{-1}x-cos^{-1}y)\) as algebraic in x and y as x \(\sqrt{1-y^{2}}+y\sqrt{1-x^{2}}\)