Evaluate exactly.
-$\sec \left( \tan ^ { - 1 } \left( \frac { - 3 } { 4 } \right) \right)$
A) $\frac { 5 } { 4 }$
B) $- \frac { 5 } { 3 }$
C) $\frac { 1 } { 4 }$
D) $- \frac { 5 } { 4 }$

Question

Quizplus · Accepted Answer

We need to evaluate $\sec(	an^{-1}(-3/4))$. Let $	heta = 	an^{-1}(-3/4)$. Then $	an	heta = -\frac{3}{4}$ and we draw a right triangle with angle $	heta$ such that the opposite side has length $-3$ and the adjacent side has length $4$ (since $	an	heta = \frac{	ext{opposite}}{	ext{adjacent}}$).

[asy]
pair A,B,C;
A = (0,0);
B = (4,0);
C = (4,-3);
draw(rightanglemark(B,A,C,5));
draw(A--B--C--A);
draw(C--B,red);
label("$4$", (A+B)/2, N);
label("$-3$", (C+A)/2, W);
label("$5$", (C+B)/2, SE);
label("$	heta$", A + (0.4,-0.2));
[/asy]

Then by the Pythagorean theorem, the hypotenuse has length $\sqrt{4^2 + (-3)^2} = 5$. Therefore, $\cos	heta = \frac{	ext{adjacent}}{	ext{hypotenuse}} = \frac{4}{5}$ and $\sec	heta = \frac{1}{\cos	heta} = \boxed{\frac{5}{4}}$.

Evaluate Exactly sec⁡(tan⁡−1(−34))\sec \left( \tan ^ { - 1 } \left( \frac { - 3 } { 4 } \right) \right)sec(tan−1(4−3​))

Evaluate Exactly $\sec \left( \tan ^ { - 1 } \left( \frac { - 3 } { 4 } \right) \right)$