Finding Tangent of Angle with Known Sine
The expression given is:
\[ \tan(\sin^{-1}(\frac{\sqrt{2}}{2})) \]
Here, we are asked to find the tangent of the angle whose sine is \(\frac{\sqrt{2}}{2}\).
Recognize that \(\sin^{-1}(\frac{\sqrt{2}}{2})\) corresponds to the angle whose sine is \(\frac{\sqrt{2}}{2}\). This is an angle in a right triangle where the side opposite the angle has length \(\sqrt{2}\) and the hypotenuse has length 2. By the Pythagorean theorem, the length of the adjacent side of this right triangle will be:
\[ \text{Adjacent side} = \sqrt{\text{Hypotenuse}^2 - \text{Opposite side}^2} \]
\[ \text{Adjacent side} = \sqrt{2^2 - (\sqrt{2})^2} \]
\[ \text{Adjacent side} = \sqrt{4 - 2} \]
\[ \text{Adjacent side} = \sqrt{2} \]
Now we have a right triangle where the opposite side is \(\sqrt{2}\), the adjacent side is also \(\sqrt{2}\), and the hypotenuse is 2. To find the tangent of the angle, we simply divide the opposite side by the adjacent side:
\[ \tan(\sin^{-1}(\frac{\sqrt{2}}{2})) = \frac{\text{Opposite side}}{\text{Adjacent side}} \]
\[ \tan(\sin^{-1}(\frac{\sqrt{2}}{2})) = \frac{\sqrt{2}}{\sqrt{2}} \]
\[ \tan(\sin^{-1}(\frac{\sqrt{2}}{2})) = 1 \]
Therefore, the value of the expression is 1.
