Question

Assume that 0 < x < pi/2 and
0 < y < pi/2. Find the exact value of tan(x + y) if csc x = 5/3 and cos y = 5/13

Answer 1

The Answer is D. -63/16

Answer 2

The exact value of $\tan (x+y)$ is $-\frac{63}{16}$.

To find the exact value of $\tan (x+y)$, we shall use the following trigonometric identities:

$\csc x = \frac{1}{\sin x}$ (1)

$\tan (x+y) = \frac{\tan x + \tan y}{1-\tan x\cdot \tan y}$ (2)

$\tan x = \frac{\sin x}{\cos x}$ (3)

$\tan y = \frac{\sin y}{\cos y}$ (4)

$\cos x = \sqrt{1-\sin^{2}x}$ (5)

$\sin y = \sqrt{1-\cos^{2}y}$ (6)

By (1), we find that the sine of $x$ is:

$\sin x = \frac{3}{5}$

And the remaining trigonometric identities are determined below:

$\cos x = \sqrt{1-\left(\frac{3}{5} \right)^{2}}$

$\cos x = \frac{4}{5}$

$\sin y = \sqrt{1-\left(\frac{5}{13} \right)^{2}}$

$\sin y = \frac{12}{13}$

Lastly, we determine the exact value:

$\tan (x+y) = \frac{\tan x + \tan y}{1-\tan x\cdot \tan y}$

$\tan (x+y) = \frac{\frac{\sin x}{\cos x} + \frac{\sin y}{\cos y} }{1 -\left(\frac{\sin x}{\cos x} \right)\cdot \left(\frac{\sin y}{\cos y} \right)}$

$\tan (x+y) = \frac{\frac{3}{4}+\frac{12}{5}}{1-\left(\frac{3}{4} \right)\cdot \left(\frac{12}{5} \right)}$

$\tan (x+y) = -\frac{63}{16}$

We kindly invite to check this question on trigonometric identities: https://brainly.com/question/24377281