Why not multiple both denominators by 1 - sin(x)?I think in order to solve this I need to get [MATH]sin^2x+cos^3x[/MATH] in numorator so I can divide it by cosx (1+sinx) and get 1/(1+sinx) or smth like that just tried like 20 different ways using all sorts of identities and cant get there
Start with RHS:[MATH]1/(1+sinx)=(secx-sinxsecx)/cosx[/MATH] what I am doing is multiplying right side both numerator and denominator by 1+sinx and try to go from there but cant get 1 in numerator tried so many different ways
For me it just makes the right side much more complicated. How would u solve that?Why not multiple both denominators by 1 - sin(x)?
Maybe factor the RHS numerator.
Oh okay and then multiply right side by 1-sinx which makes it 1-sinx/cos^2x think I got it now that really helps thnxStart with RHS:
\(\displaystyle \displaystyle{\dfrac{sec(x) - sin(x) * sec(x)}{cos(x)}}\)
\(\displaystyle = \ \displaystyle{\dfrac{sec(x) * [1- sin(x)]}{cos(x)}}\)
\(\displaystyle = \ \displaystyle{\dfrac{ 1- sin(x)}{cos^2(x)}}\)
\(\displaystyle = \ \displaystyle{\dfrac{ 1 \ - \ sin(x)}{1 \ - \ sin^2(x)}}\)
continue......
But how did you get from - sinx*secx to 1-sinx?Start with RHS:
\(\displaystyle \displaystyle{\dfrac{sec(x) - sin(x) * sec(x)}{cos(x)}}\)
\(\displaystyle = \ \displaystyle{\dfrac{sec(x) * [1- sin(x)]}{cos(x)}}\)
\(\displaystyle = \ \displaystyle{\dfrac{ 1- sin(x)}{cos^2(x)}}\)
\(\displaystyle = \ \displaystyle{\dfrac{ 1 \ - \ sin(x)}{1 \ - \ sin^2(x)}}\)
continue......
But how did you get from - sinx*secx to 1-sinx?
Are you just staring at the screen or working with pencil-paper?But how did you get from - sinx*secx to 1-sinx?
\(\displaystyle sec(x) = \frac{1}{cos(x)}\)
By practicing, with a wide variety of exercises. After sufficient exposure, things become easier.... like how am I suppose to know that? ...