## Precalculus (6th Edition) Blitzer

The required value of the expression is $\frac{xy+\sqrt{1-{{y}^{2}}}\sqrt{1-{{x}^{2}}}}{y\sqrt{1-{{x}^{2}}}-x\sqrt{1-{{y}^{2}}}}.$
Step 1: $\sin \theta =\frac{\text{Perpendicular}\left( \text{P} \right)}{\text{Hypotenuse}\left( \text{H} \right)}$ $\cos \theta =\frac{\text{Base}\left( \text{B} \right)}{\text{Hypotenuse}\left( \text{H} \right)}$ $\tan \theta =\frac{\text{Perpendicular}\left( \text{P} \right)}{\text{Base}\left( \text{B} \right)}$ Let us suppose ${{\sin }^{-1}}x=\alpha$ \begin{align} & x=\sin \alpha \\ & \sin \alpha =\frac{x}{1} \\ & =\frac{\text{P}}{\text{H}} \end{align} By using the Pythagorian Theorem, ${{\left( \text{Hypotenuse} \right)}^{2}}={{\left( \text{Perpendicular} \right)}^{2}}+{{\left( \text{Base} \right)}^{2}}$ \begin{align} & {{\text{H}}^{2}}={{\text{P}}^{2}}+{{\text{B}}^{2}} \\ & {{1}^{2}}={{x}^{2}}+{{\text{B}}^{2}} \\ & {{\text{B}}^{2}}=1-{{x}^{2}} \\ & \text{B}=\sqrt{1-{{x}^{2}}} \\ \end{align} Then, \begin{align} & \tan \alpha =\frac{\text{P}}{\text{B}} \\ & =\frac{x}{\sqrt{1-{{x}^{2}}}} \end{align} Step 2: Similarly, ${{\cos }^{-1}}y=\beta$. This implies, \begin{align} & y=\cos \beta \\ & \cos \beta =\frac{y}{1} \\ & =\frac{\text{B}}{\text{H}} \end{align} By using the Pythagorian Theorem, ${{\left( \text{Hypotenuse} \right)}^{2}}={{\left( \text{Perpendicular} \right)}^{2}}+{{\left( \text{Base} \right)}^{2}}$ \begin{align} & {{\text{H}}^{2}}={{\text{P}}^{2}}+{{\text{B}}^{2}} \\ & {{1}^{2}}={{\text{P}}^{2}}+{{y}^{2}} \\ & {{\text{P}}^{2}}=1-{{y}^{2}} \\ & \text{P}=\sqrt{1-{{y}^{2}}} \\ \end{align} So, \begin{align} & \tan \beta =\frac{\text{P}}{\text{B}} \\ & =\frac{\sqrt{1-{{y}^{2}}}}{y} \end{align} Step 3: Using step 1 and step 2, $\tan \left( {{\sin }^{-1}}x+{{\cos }^{-1}}y \right)$ can be written as $\tan \left( {{\sin }^{-1}}x+{{\cos }^{-1}}y \right)=\tan \left( \alpha +\beta \right)$ By using the trigonometric identity, $\tan \left( \alpha +\beta \right)=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$ From above steps, $\tan \alpha =\frac{x}{\sqrt{1-{{x}^{2}}}}$ , $tan\beta =\frac{\sqrt{1-{{y}^{2}}}}{y}$ Thus, \begin{align} & \tan \left( \alpha +\beta \right)=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta } \\ & =\frac{\frac{x}{\sqrt{1-{{x}^{2}}}}+\frac{\sqrt{1-{{y}^{2}}}}{y}}{1-\left( \frac{x}{\sqrt{1-{{x}^{2}}}}\times \frac{\sqrt{1-{{y}^{2}}}}{y} \right)} \\ & =\frac{\frac{xy+\sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}}}}{y\sqrt{1-{{x}^{2}}}}}{1-\frac{x\sqrt{1-{{y}^{2}}}}{y\sqrt{1-{{x}^{2}}}}} \\ & =\frac{\frac{xy+\sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}}}}{y\sqrt{1-{{x}^{2}}}}}{\frac{y\sqrt{1-{{x}^{2}}}-x\sqrt{1-{{y}^{2}}}}{y\sqrt{1-{{x}^{2}}}}}=\frac{xy+\sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}}}}{y\sqrt{1-{{x}^{2}}}-x\sqrt{1-{{y}^{2}}}} \end{align}