Verify the identity

[PanTh3R]

Verify the Identity:

\(\displaystyle \frac{\sec{\theta}-1}{1-\cos{\theta}} = \sec\theta\)

i did... \(\displaystyle \frac{\sec{\theta}-1}{1-\cos{\theta}}*\frac{1+\cos{\theta}}{1+\cos{\theta}}\)

\(\displaystyle \frac{(\sec\theta-1)(1+\cos\theta)}{1-\cos^2\theta}\)


\(\displaystyle \frac{\sec\theta+\sec\theta\cos\theta-1-\cos\theta}{\sin^2\theta}\)

and now im stuck i think i did the first steps right...can any one help please?

 

[mmm4444bot]

Verify the Identity:

\(\displaystyle \frac{\sec{\theta}-1}{1-\cos{\theta}} = \cos\theta\) This is not an identity.

Did you make a typographical error?

We can get an identity, if we change the righthand side to the secant of theta.

\(\displaystyle \frac{\sec{\theta}-1}{1-\cos{\theta}} = \sec{\theta}\)

You arrived at the following.

\(\displaystyle \frac{\sec\theta+\sec\theta\cos\theta-1-\cos\theta}{\sin^2\theta}\)

As soon as we think of sec(?) as 1/cos(?), it's an easy step to simplify sec(?) cos(?) - 1, in the numerator above.

Then subtract cos(?) from sec(?), in the resulting numerator. From there, it's just a compound ratio simplification.

The entire verification of this identity is about three steps, if the first step is restating the secant functions in terms of cosine.

\(\displaystyle \frac{\sec{\theta}-1}{1-\cos{\theta}} = \sec{\theta}\)

\(\displaystyle \frac{\frac{1}{cos\theta} - \frac{cos\theta}{cos\theta}}{1 - cos\theta} = \frac{1}{cos\theta}\)

\(\displaystyle \frac{\frac{1 - cos\theta}{cos\theta}}{1 - cos\theta} = \frac{1}{cos\theta}\)

\(\displaystyle \frac{1 - cos\theta}{cos\theta} \cdot \frac{1}{1 - cos\theta} = \frac{1}{cos\theta}\)

:idea: Several identities that involve secant, cosecant, and sometimes tangent and cotangent, can be verified more easily by first restating everything in terms of sine and cosine.

 

[soroban]

Hello, PanTh3R!

mmm444bot is right . . . There is a typo.

\(\displaystyle \text{Verify the Identity: }\;\frac{\sec\theta - 1}{1 - \cos\theta} \;=\;\boxed{\sec\theta}\)

\(\displaystyle \text{We have: }\;\frac{\sec\theta - 1}{1 - \frac{1}{\sec\theta}}\)

\(\displaystyle \text{Multiply by }\frac{\sec\theta}{\sec\theta}\!:\;\;\frac{\sec\theta(\sec\theta - 1)}{\sec\theta\left(1 - \frac{1}{\sec\theta}\right)} \;=\;\frac{\sec\theta(\sec\theta -1)}{\sec\theta - 1} \;=\; \sec\theta\)

 

[PanTh3R]

wow i did make a typo error it was sec(theta) thanks for helping me :mrgreen: