## anonymous 4 years ago (1+SINX)^(1/2) 1+sinx -------------= ------- (1-sinx)^(1/2) |cosx|

1. anonymous

verify

2. anonymous

okay, you have what exactly? this? $\sqrt{\frac{1+sinx}{1-sin(x)}}=\frac{1+sin(x)}{|cos(x)|}\text{?}$

3. anonymous

correct

4. dumbcow

multiply top and bottom by conjugate: sqrt(1+sin x) $\frac{\sqrt{1+sinx}}{\sqrt{1-sinx}}*\frac{\sqrt{1+sinx}}{\sqrt{1+sinx}} = \frac{1+sinx}{\sqrt{1-\sin^{2} x}} = \frac{1+sinx}{cosx}$

5. campbell_st

rationalise the denominator by multiplying by 1 + sin(x) will result in $\sqrt{(1+\sin(x)^2/1 - \sin^2(x)}$ denominator is cos^2(x) while square root of the numerator is (1+sin(x)

6. anonymous

after the sqrt gets off the top wouldn't it go off the bottom as well

7. campbell_st

the reason its absolute value is that the square root will give a positive and negative answer.. hence the abs value