Question

If f(x)=1-x/1+x then find f[f(sin theta)]

Answer 1

Answer:

$f[f(sin\theta)]=sin\theta$

Step-by-step explanation:

Given:

$f(x)=\frac{1-x}{1+x}$

Now,

$f(sin\theta)$

$=\frac{1-sin\theta}{1+sin\theta}$

$f[f(sin\theta)]$

$=f[\frac{1-sin\theta}{1+sin\theta}]$

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

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

$=\frac{1+sin\theta-1+sin\theta}{1+sin\theta+1-sin\theta}$

$=\frac{sin\theta+sin\theta}{1+1}$

$=\frac{2sin\theta}{2}$

$=sin\theta$

we get

$f[f(sin\theta)]=sin\theta$