# Verifying trigonometric identities?

• Feb 4th 2011, 07:19 PM
homeylova223
Verifying trigonometric identities?
Can anyone help me verify these two trigonometric identities

(sinA+cosA)^2= 2+ secAcscA
___________
secAcscA
2/(secAcscA)+ (secAcscA/secAcscA)
2/(secAcscA)+1
2(1/sec+1/csc)+1
2cosAsinA+1
Unfortunately I am stuck and do not know how to proceed(Worried)

For my second problem
sinx+cosx= (cosx/1-tanx)+(sinx/1-cot)
(cosx/(1-sin/cos)+(sinx/(1-cosx/sinx)
I am unsure how to proceed

• Feb 4th 2011, 07:31 PM
Prove It
2.

$\displaystyle \frac{\cos{x}}{1 - \tan{x}} + \frac{\sin{x}}{1 - \cot{x}} = \frac{\cos{x}}{1 - \frac{\sin{x}}{\cos{x}}} + \frac{\sin{x}}{1 - \frac{\cos{x}}{\sin{x}}}$

$\displaystyle = \frac{\cos{x}}{\frac{\cos{x} - \sin{x}}{\cos{x}}} + \frac{\sin{x}}{\frac{\sin{x} - \cos{x}}{\sin{x}}}$

$\displaystyle = \frac{\cos^2{x}}{\cos{x} - \sin{x}} + \frac{\sin^2{x}}{\sin{x} - \cos{x}}$

$\displaystyle = \frac{\cos^2{x}}{\cos{x} - \sin{x}} - \frac{\sin^2{x}}{\cos{x} - \sin{x}}$

$\displaystyle = \frac{\cos^2{x} - \sin^2{x}}{\cos{x} - \sin{x}}$

$\displaystyle = \frac{(\cos{x} - \sin{x})(\cos{x} + \sin{x})}{\cos{x} - \sin{x}}$

$\displaystyle = \cos{x} + \sin{x}$.
• Feb 4th 2011, 07:54 PM
Soroban
Hello, homeylova223!

Quote:

$(\sin A+\cos A)^2 \:=\:\dfrac{2+ \sec A\csc A}{\sec A\csc A}$

$(\sin A + \cos A)^2 \;=\;\sin^2\!A + 2\sin A\cos A + \cos^2\!A$

. . . . . . . . . . . . $=\; 2\sin A\cos A + \underbrace{\sin^2\!A + \cos^2\!A}}_{\text{This is 1}}$
. . . . . . . . . . . . $=\;2\sin A\cos A + 1$

. . . . . . . . . . . . $=\;\dfrac{2}{\sec A\csc A} + 1$

. . . . . . . . . . . . $=\;\dfrac{2 + \sec A\csc A}{\sec A\csc A}$

• Feb 5th 2011, 09:38 AM
homeylova223
Quick question on number 2 on the bottom part where you have (1-sinx/cosx) times cos x
How do you get cosx-sinx/cosx I understand the rest of the problem I am just slightly confused on that part.
• Feb 5th 2011, 05:13 PM
Prove It
Quote:

Originally Posted by homeylova223
Quick question on number 2 on the bottom part where you have (1-sinx/cosx) times cos x
How do you get cosx-sinx/cosx I understand the rest of the problem I am just slightly confused on that part.

Sorry but without brackets in the correct spots I can't understand what you're writing...
• Feb 5th 2011, 07:22 PM
homeylova223
Like when you have
(cosx/ 1-sinx/cosx) how do you get I

(cosx/(cosx-sinx/cosx)

I am slightly confused on that part
• Feb 5th 2011, 07:26 PM
harish21
Do you mean $\frac{\cos(x)}{1-\frac{\sin(x)}{\cos(x)}} = \frac{\cos(x)}{\frac{\cos(x)-\sin(x)}{\cos(x)}}$???

Note that:

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

Therefore, $\dfrac{\cos(x)}{1-\dfrac{\sin(x)}{\cos(x)}} = \dfrac{\cos(x)}{\dfrac{\cos(x)-\sin(x)}{\cos(x)}}$
• Feb 6th 2011, 12:02 PM
homeylova223
Right now I have two more problems that are similar to the ones I just did can anyone help me with these?

It is like this

(1+tanx)/(sinx+cosx)= secx
(secx)/(1)=secx
(1/cosx/1)= secx
secx=secx
Did I do this right?

My second problem is
sinx+cosx= (2 sin^2x-1)/(sinx-cosx) I am not sure how to this one
• Feb 6th 2011, 12:12 PM
skeeter
Quote:

Originally Posted by homeylova223
Right now I have two more problems that are similar to the ones I just did can anyone help me with these?

It is like this

(1+tanx)/(sinx+cosx)= secx
(secx)/(1)=secx ... no, 1+tanx is not secx, 1 + tan^2x = sec^2x. also sinx+cosx is not 1, it's sin^2x+cos^2x = 1
(1/cosx/1)= secx
secx=secx
Did I do this right?

My second problem is
sinx+cosx= (2 sin^2x-1)/(sinx-cosx) I am not sure how to this one

on the RHS, numerator ... 2sin^2x - 1 = 2sin^2x - (sin^2x + cos^2x) = sin^2x - cos^2x ... which will factor.

...
• Feb 6th 2011, 12:26 PM
homeylova223
So on
(1+tanx)/(sinx+cosx)= secx

Would I do it like this

(1/sinx+cosx)+(sinx/cosx)/(sinx+cosx)= secx
Then I am not sure what to do?
• Feb 6th 2011, 01:01 PM
skeeter
Quote:

Originally Posted by homeylova223
So on
(1+tanx)/(sinx+cosx)= secx

Would I do it like this

(1/sinx+cosx)+(sinx/cosx)/(sinx+cosx)= secx
Then I am not sure what to do?

on the left side ...

$\dfrac{1+\tan{x}}{\sin{x}+\cos{x}} \cdot \dfrac{\cos{x}}{\cos{x}} = \dfrac{\cos{x}+\sin{x}}{(\sin{x}+\cos{x})\cos{x}}$

finish it