Question

Help me. It's 3a.m. in the morning and I'm stuck on my last question.

Prove: cos^2x-sin^2x/cosx+sinx = cscx - secx/secx cscx

Answer 1

\( \dfrac{\cos^2x-\sin^2x}{\cos x+\sin x} = \dfrac{\csc x - \sec x}{\sec x \csc x} \)

Is this it?

Answer 2

yea

Answer 3

I've tried both sides but it doesn't seem to work out...

Answer 4

well why not factor the numerator

\[\frac{(\cos(x) + \sin(x))(\cos(x) - \sin(x))}{\sin(x) + \cos(x)}\]

Answer 5

doesn't that just leave cosx-sinx

Answer 6

well use the fact that the numerator is

cos(2x)

and

cos(2x) = 1 - 2sin^2(x)

or

cos(2x) - 2cos^2(x) - 1

or

cos(2x) = (1 - tan^2(x))/(1 + tan^2(x))

Answer 7

\(\large \dfrac{\cos^2x-\sin^2x}{\cos x+\sin x} = \dfrac{\csc x - \sec x}{\sec x \csc x}\)

\(\large = \dfrac{\frac{1}{\sin x} -\frac{1}{\cos x} } {\frac{1}{\cos x}{\frac{1}{\sin x} }} \)

\(\large = \dfrac{\frac{1}{\sin x} -\frac{1}{\cos x} } {\frac{1}{\cos x}{\frac{1}{\sin x} }} \times \dfrac{\sin x \cos x}{\sin x \cos x} \)

\(\large = \dfrac{\cos x -\sin x} {1} \)

\(\large = \dfrac{\cos x -\sin x} {1} \times \dfrac{\cos x + \sin x}{\cos x + \sin x} \)

\(\large = \dfrac{\cos^2 x -\sin^2 x} {\cos x + \sin x} \)

Answer 8

I have another question, is the problem: (secx+cscx)cotx = false? because everyone cant seem to solve it...

Answer 9

You can certainly multiply that out. Is it supposed to be equal to something?

Answer 10

so you have

1/cos * cos/sin + 1/sin * cos/sin = 1/sin + cos/sin^2

so csc + csc*tan

which becomes

csc(1 + tan)

Answer 11

oh it's supposed to be equal to 1

Answer 12

Hmm these are all good methods :)
This is how I would have done it though,\[\Large\bf\sf \frac{(\cos x-\sin x)(\cos x+\sin x)}{\cos x+ \sin x}\quad=\quad \frac{\csc x - \sec x}{\sec x \csc x}\]
\[\Large\bf\sf \cos x - \sin x quad=\quad \frac{\csc x - \sec x}{\sec x \csc x}\]
\[\Large\bf\sf \frac{1}{\sec x}-\frac{1}{\csc x} \quad=\quad \frac{\csc x - \sec x}{\sec x \csc x}\]Then just get a common denominator in the left side.

Answer 13

Grr stupid latex fail t.t

Answer 14

\( (\sec x + \csc x) \cot x~~~~~~~~~~~ = 1 \)

\(\left( \dfrac{1}{\cos x} + \dfrac{1}{\sin x} \right) \dfrac{\cos x}{\sin x}~~~~ = 1\)

\( \dfrac{1}{\sin x} + \dfrac{\cos x}{\sin^2 x} ~~~~~~~~~~~~~~~~~=1\)

\( \dfrac{1}{\sin x} \times \dfrac{\sin x}{\sin x} + \dfrac{\cos x}{\sin^2 x} ~~~~~=1\)

\(\dfrac{\sin x}{\sin^2 x} + \dfrac{\cos x}{\sin^2 x} ~~~~~~~~~~~~~~~~=1\)

\(\dfrac{\sin x + \cos x}{\sin^2 x} ~~~~~~~~~~~~~~~~~~~~=1\)

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

It doesn't look like an identity to me.
This may be true for certain values of x which makes this an equation, not an identity.

Answer 15

Yea, it's okay, I knew it wouldn't work...But thanks a ton. I wish I could hug you and throw you some food but I can't, sadly... >n<

Answer 16

LOL. You're welcome.