# Verify Identity

#### powp

Hello All

I am having problems Verifying this identity

(COS A) / (1-SIN A) = SEC A + TAN A

I can get the RHS = (1 + SIN A)/COS A

But this does not equal the LHS

THanks

#### quetzalcoatl9

yes, you are right, now cross-multiply:

$$cos^2 a = (1 + sin a) (1 - sin a)$$
$$cos^2 a = 1 - sin^2 a$$
$$sin^2 a + cos^2 a = 1$$

#### JasonJo

sec A + tan A = (1/cosA) + sinA/cosA = (cosA + sinA*cosA)/(cosA)^2 =

cosA(1 + sinA)/(1-sin^2A) = cos A / (1 + sinA)

#### powp

does $$sin^2 a + cos^2 a = 1$$ verify the idenity?

#### James R

Homework Helper
Gold Member
Divide the LHS top and bottom by cos A:

$$\frac{\cos A}{1-\sin A} = \frac{1}{\sec A - \tan A}$$
$$=\frac{1}{(\sec A - \tan A)}\frac{(\sec A + \tan A)}{(\sec A + \tan A)}$$
$$=\frac{\sec A + \tan A}{\sec^2 A - \tan^2 A}$$

But

$$\sec^2 A - \tan^2 A = \frac{1 - \sin^2 A}{\cos^2 A} = 1$$

So, the result follows.

#### HallsofIvy

Homework Helper
powp said:
Hello All

I am having problems Verifying this identity

(COS A) / (1-SIN A) = SEC A + TAN A

I can get the RHS = (1 + SIN A)/COS A

But this does not equal the LHS

THanks
Yes, it does: (1+ sin A)/cos A= 1/cos A+ sin A/cos A= sec A+ tan A.

#### powp

Thanks All Still kind of confused.

HallsofIvy: doesn't that just change back the RHS that I did??

#### James R

Homework Helper
Gold Member
powp:

Does this help?

$$\frac{1 + \sin A}{\cos A} = \frac{(1 + \sin A)(1 - \sin A)}{\cos A(1 - \sin A)}$$

$$=\frac{1 - \sin^2 A}{\cos A (1 - \sin A)}$$

$$= \frac{\cos^2 A}{\cos A (1 - \sin A)}$$

$$= \frac{\cos A}{1- \sin A}$$

#### uart

powp said:
does $$sin^2 a + cos^2 a = 1$$ verify the idenity?
Yes that is perhaps the most fundemental of all trig identities. It's actually just Pythagoras Thm for a RHT with unit length hypotemus.

#### powp

Is this correct??

Hello All

I need to find all solutions to the following did i do it correct?

Sin 2x = 2Tan 2x

2Sinx Cosx = 2(2Tanx / 1-Tan^2x)

Sinx Cosx = (2(Sinx / Cosx)/(cos^2x-Sin^2x / Cos^2x))

Sinx Cosx = 2(Sinx / Cosx) X (Cos^2x / cos^2x-Sin^2x)

Cosx = 2 (Cosx / Cos^2x - Sin^2x)

1 = (2 / Cos^2x - Sin^2x)

Cos^2x - Sin^2x = 2

1 - Sinx^2 - Sin^2x = 2

2Sin^2x = -1
Sin^2x = -1/2

Sinx = -1/SQROOT(2)

Is this correct??

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