# Thread: I need help asap i would appreciate it if anybody could help me thanks!!!!

1. ## I need help asap i would appreciate it if anybody could help me thanks!!!!

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

2. Originally Posted by angelica2619
I need help with these problems....

1. Is cosX=1/4 and 0<x<90, find cscx

[snip]
1. $\csc{x} = \frac{1}{\sin{x}}$

Remember the Pythagorean Identity...

$\cos^2{x} + \sin^2{x} = 1$

And since $\cos{x} = \frac{1}{4}$

$\left(\frac{1}{4}\right)^2 + \sin^2{x} = 1$

$\frac{1}{16} + \sin^2{x} = 1$

$\sin^2{x} = \frac{15}{16}$

$\sin{x} = \sqrt{\frac{15}{16}}$

$\sin{x} = \frac{\sqrt{15}}{4}$

$\frac{1}{\sin{x}} = \frac{4}{\sqrt{15}}$

$\frac{1}{\sin{x}} = \frac{4\sqrt{15}}{15}$

$\csc{x} = \frac{4\sqrt{15}}{15}$.

Originally Posted by angelica2619
[snip]
2. If cotx=square root of -6/3, and 90<x<180, find tan

[snip]

2. You can't have the square root of a negative number.

3. Originally Posted by angelica2619
[snip]
3.Simplify tanx/sinx

[snip]
3. Remember that $\tan{x} = \frac{\sin{x}}{\cos{x}}$.

So $\tan{x} \div \sin{x} = \frac{\sin{x}}{\cos{x}} \div \sin{x}$

$= \frac{\sin{x}}{\cos{x}} \times \frac{1}{\sin{x}}$

$= \frac{\sin{x}}{\sin{x}\cos{x}}$

$= \frac{1}{\cos{x}}$

$= \sec{x}$.

4. Originally Posted by angelica2619
[snip]
4.simplify cotxtanx+sinxsecx

[snip]
4. Remember that $\cot{x} = \frac{1}{\tan{x}}$ and $\sec{x} = \frac{1}{\cos{x}}$.

So $\cot{x}\tan{x} = \frac{1}{\tan{x}}\tan{x} = 1$

and $\sin{x}\sec{x} = \sin{x}\frac{1}{\cos{x}} = \tan{x}$.

Therefore $\cot{x}\tan{x} + \sin{x}\sec{x} = 1 + \tan{x}$.