I need help with these problems....
1. Is cosX=1/4 and 0<x<90, find cscx
2. If cotx=square root of -6/3, and 90<x<180, find tan
3.Simplify tanx/sinx
4.simplify cotxtanx+sinxsecx
I need help with these problems....
1. Is cosX=1/4 and 0<x<90, find cscx
2. If cotx=square root of -6/3, and 90<x<180, find tan
3.Simplify tanx/sinx
4.simplify cotxtanx+sinxsecx
1. $\displaystyle \csc{x} = \frac{1}{\sin{x}}$
Remember the Pythagorean Identity...
$\displaystyle \cos^2{x} + \sin^2{x} = 1$
And since $\displaystyle \cos{x} = \frac{1}{4}$
$\displaystyle \left(\frac{1}{4}\right)^2 + \sin^2{x} = 1$
$\displaystyle \frac{1}{16} + \sin^2{x} = 1$
$\displaystyle \sin^2{x} = \frac{15}{16}$
$\displaystyle \sin{x} = \sqrt{\frac{15}{16}}$
$\displaystyle \sin{x} = \frac{\sqrt{15}}{4}$
$\displaystyle \frac{1}{\sin{x}} = \frac{4}{\sqrt{15}}$
$\displaystyle \frac{1}{\sin{x}} = \frac{4\sqrt{15}}{15}$
$\displaystyle \csc{x} = \frac{4\sqrt{15}}{15}$.
2. You can't have the square root of a negative number.
3. Remember that $\displaystyle \tan{x} = \frac{\sin{x}}{\cos{x}}$.
So $\displaystyle \tan{x} \div \sin{x} = \frac{\sin{x}}{\cos{x}} \div \sin{x}$
$\displaystyle = \frac{\sin{x}}{\cos{x}} \times \frac{1}{\sin{x}}$
$\displaystyle = \frac{\sin{x}}{\sin{x}\cos{x}}$
$\displaystyle = \frac{1}{\cos{x}}$
$\displaystyle = \sec{x}$.
4. Remember that $\displaystyle \cot{x} = \frac{1}{\tan{x}}$ and $\displaystyle \sec{x} = \frac{1}{\cos{x}}$.
So $\displaystyle \cot{x}\tan{x} = \frac{1}{\tan{x}}\tan{x} = 1$
and $\displaystyle \sin{x}\sec{x} = \sin{x}\frac{1}{\cos{x}} = \tan{x}$.
Therefore $\displaystyle \cot{x}\tan{x} + \sin{x}\sec{x} = 1 + \tan{x}$.