91Ó°ÊÓ

Using the basic trigonometric identities, we can write the tangent function in terms of the sine and cosine functions: \[ \tan(x) = \frac{\sin(x)}{\cos(x)} \]

By definition, if \(y = \cos^{-1}(t)\), then \[ \cos(y) = t \] with \(0 < y \leq \pi\). Now, we need to relate the tangent function with this inverse cosine function.

Now, let's consider the triangle representation of the trigonometric functions. Let's have a right-angled triangle with one angle equals to \(y\), and the opposite side \(a\), adjacent side \(b\), and the hypotenuse side \(c\). Using the sine and cosine functions on this triangle: \[ \sin(y)=\frac{a}{c} \quad \text{and} \quad \cos(y)=\frac{b}{c} \] Now, given that \(t = \cos(y)\): \[ c = \frac{b}{t} \] As we know, the Pythagorean theorem states that \(a^2 + b^2 = c^2\). Substituting the value of \(c\) into this equation: \[ a^2 + b^2 = \left(\frac{b}{t}\right)^2 \] Solving for \(a^2\): \[ a^2 = \frac{b^2}{t^2} - b^2 = b^2 \left(\frac{1}{t^2} - 1\right) \] Now, using the sine and cosine functions to find the tangent function: \[ \tan(y) = \frac{\sin(y)}{\cos(y)} = \frac{a}{b} \] So, from our previous derived equation, we have: \[ \tan(y) = \frac{a}{b} = \frac{\sqrt{b^2\left(\frac{1}{t^2} - 1\right)}}{b} \] Finally, we can simplify this expression by dividing the numerator and denominator by \(b\): \[ \tan(y) = \frac{\sqrt{\frac{1}{t^2} - 1}}{t} \]

Now we apply the inverse tangent function on both sides: \[ \tan^{-1}\left(\tan(y)\right) = \tan^{-1}\left(\frac{\sqrt{1 - t^2}}{t}\right) \] By definition, since \(y = \cos^{-1}(t)\), we can now write the given equation as: \[ \cos^{-1}(t) = \tan^{-1}\left(\frac{\sqrt{1 - t^2}}{t}\right) \] Hence, the given equation holds true for \(0 < t \leq 1\).