Question

Evaluate the expression without using a calculator. (Hint: Make a sketch of a right triangle, as illustrated in Example \(7 .)\) (a) \(\sec \left[\arctan \left(-\frac{3}{5}\right)\right]\) (b) \(\tan \left[\arcsin \left(-\frac{5}{6}\right)\right]\)

Short answer

(a) \(\sec \left[\arctan \left(-\frac{3}{5}\right)\right] = \frac{\sqrt{34}}{5}\)\n(b) \(\tan \left[\arcsin \left(-\frac{5}{6}\right)\right] = \frac{-5}{\sqrt{11}}\)

Step-by-step solution

Evaluate (a)

To evaluate \(\sec \left[\arctan \left(-\frac{3}{5}\right)\right]\), starting by interpreting \(\arctan \left(-\frac{3}{5}\right)\) as the angle in a right-angled triangle whose opposite side is -3 and adjacent side is 5. The hypotenuse of this triangle can be calculated using the Pythagorean theorem \(a^2 + b^2 = h^2\), giving \(\sqrt{(-3)^2 + 5^2} = \sqrt{9+25} = \sqrt{34}\). Therefore, the secant of that angle is the ratio of the hypotenuse to the adjacent side, giving \(\sec \left[\arctan \left(-\frac{3}{5}\right)\right] = \frac{\sqrt{34}}{5}\).

Evaluate (b)

To evaluate \(\tan \left[\arcsin \left(-\frac{5}{6}\right)\right]\), interpret \(\arcsin \left(-\frac{5}{6}\right)\) as the angle in a right-angled triangle whose opposite side is -5 and hypotenuse is 6. The adjacent side of this triangle can be calculated using the Pythagorean theorem as \(\sqrt{6^2 - (-5)^2} = \sqrt{36 - 25} = \sqrt{11}\). Hence, the tangent of this angle is the ratio of the of the opposite side to the adjacent side that is \(\tan \left[\arcsin \left(-\frac{5}{6}\right)\right] = \frac{-5}{\sqrt{11}}\).

Simplify (b)

Avoid rationalizing the denominator because it is unnecessary in modern mathematics and leads to an equivalent answer, therefore, \(\tan \left[\arcsin \left(-\frac{5}{6}\right)\right] = \frac{-5}{\sqrt{11}}\) is the most simplified form.