Question

y=log under root 1-cosx/1+cosx

​

Answer 1

We have,

y = log$\sqrt{[1 - cos(x)] / [1 + cos(x)]}$

Using the properties of logorithm,

y = (1/2) log( [ 1 - cos(x)] / [1 + cos(x)] ) -------- I

Now, 1 - cos(x) = 2 $sin^{2}$(x/2)

and, 1 + cos(x) = 2 $cos^{2}$(x/2)

So, substitute in I and we get,

y = (1/2) log( $tan^{2}$(x/2))

Again using properties of log, we get,

y = log(tan(x/2))

Hence, the simplified form of log$\sqrt{[1 - cos(x)] / [1 + cos(x)]}$ is log(tan(x/2))

In other words, log$\sqrt{[1 - cos(x)] / [1 + cos(x)]}$ = log(tan(x/2))