Check out some similar questions!

tpolwort · Dec 1, 2009, 12:23 PM

I have a couple problems I cannot solve, please help me!

1. tan(x)/sec(x) + cot(x)/csc(x) =sin(x)+cos(x)

2 (1-tan(x))^2 = sec^2(x) - 2tan(x)

3. sin(x)/csc(x)-1 + sin(x)/csc(x)+1 = 2tan^2(x)

4. sin(x)[sec(x)-csc(x)] = tan(x)-

Unknown008 · Dec 1, 2009, 12:31 PM

I'll do the first one. You have to start learning your identities.

Convert all to sin and cos.

$\frac{tan(x)}{sec(x)} + \frac{cot(x)}{csc(x)} = \frac{\frac{sin(x)}{cos(x)}}{\frac{1}{cos(x)}} + \frac{\frac{cos(x)}{sin(x)}}{\frac{1}{sin(x)}}$

Then, that makes:

$\frac{\frac{sin(x)}{cos(x)}}{\frac{1}{cos(x)}} + \frac{\frac{cos(x)}{sin(x)}}{\frac{1}{sin(x)}} = \frac{sin(x)}{cos(x)}.\frac{cos(x)}{1} + \frac{cos(x)}{sin(x)}.\frac{sin(x)}{1} = sin(x) + cos(x)$

now try the others.

tpolwort · Dec 1, 2009, 12:43 PM

I have tried the others I just can't figure them out, please work me through a couple more.

Unknown008 · Dec 1, 2009, 12:46 PM

Ok, for the second one, convert tan into sin and cos. Then you expand.

You'll notice that you can convert one into 2 tan(x).

Then, convert the trig in squared into tan squared.

Then, use $tan^2(x)+1 = sec^2(x)$

Give it a try! Or say where you get stuck.

tpolwort · Dec 1, 2009, 12:49 PM

I don't know how to start the second one because it is squared on the outside of:

(1-tan(x))^2

do I make that (1-sin^2x/cos^2x) ?

Unknown008 · Dec 1, 2009, 12:52 PM

Hmm. First convert the tan.

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

and that's not $\frac{sin^2(x)}{cos^2(x)}$

Then you expand

Unknown008 · Dec 1, 2009, 12:58 PM

I have to go now. I'm leaving a last hint for the third one, then I'll be back after some 8 hours' sleep.

For the third one, combine the fractions. You'll have a difference of two squares below and use your identity to substitute that denominator.

Up, you'll have sin which you can factorise out of the expansion of your fraction (when you combine the fractions).

Then, convert all into cos and sin, and things should cross out, leaving tan^2.

tpolwort · Dec 1, 2009, 12:59 PM

I know what tan(x) equals but I am so confused as to how to start on this problem.

I figured out the third one though :)

Istill need help on the second one :/

Unknown008 · Dec 1, 2009, 01:01 PM

Expand it.

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

Convert back into tan.

Then use the identity I told you above.

tpolwort · Dec 1, 2009, 01:01 PM

Thanks so much!

Unknown008 · Dec 1, 2009, 09:53 PM

Were you able to do the last one? The last one is very easy. You just expand, convert sec and cosec into sin and cos. One will cross out, leaving 1 and the other will give tan.