Question

if secx+cosecy = 2, and sinx+cosy=k evaluate 3k/4.

Need help ASAP!

Answer 1

The equations secx+cosecy = 2, and sinx+cosy=k are trigonometry equations

The result of $\frac {3k}4$ is $\frac{3\sin(y)\cos(x)\sin(x) + 3\sin(y)\cos(x)\cos(y)}{2[\cos(x) + \sin(y)]}$

How to evaluate 3k/4

The equations are given as:

secx+cosecy = 2, and sinx+cosy=k

Divide both equations

$\frac{\sin(x) + \cos(y)}{\sec(x) + \csc(y)} = \frac k2$

Divide through by 2

$\frac{\sin(x) + \cos(y)}{2[\sec(x) + \csc(y)]} = \frac k4$

Express sec and cosec as inverse ratios

$\frac{\sin(x) + \cos(y)}{2[1/\cos(x) + 1/\sin(y)]} = \frac k4$

Take the LCM

$\frac{\sin(y)\cos(x)[\sin(x) + \cos(y)]}{2[\cos(x) + \sin(y)]} = \frac k4$

Expand

$\frac{\sin(y)\cos(x)\sin(x) + \sin(y)\cos(x)\cos(y)}{2[\cos(x) + \sin(y)]} = \frac k4$

Multiply through by 3

$\frac{3\sin(y)\cos(x)\sin(x) + 3\sin(y)\cos(x)\cos(y)}{2[\cos(x) + \sin(y)]} = \frac {3k}4$

Rewrite as:

$\frac {3k}4 =\frac{3\sin(y)\cos(x)\sin(x) + 3\sin(y)\cos(x)\cos(y)}{2[\cos(x) + \sin(y)]}$

Hence, the result of $\frac {3k}4$ is $\frac{3\sin(y)\cos(x)\sin(x) + 3\sin(y)\cos(x)\cos(y)}{2[\cos(x) + \sin(y)]}$

Read more about trigonometry equations at:

https://brainly.com/question/8120556