Question

Answer 1

\(\dfrac{sinx}{1+cosx}+\dfrac{1+cosx}{sinx}=\dfrac{sinx\left(1-cosx\right)}{\left(1+cosx\right)\left(1-cosx\right)}+\dfrac{1+cosx}{sinx}\)

\(=\dfrac{sinx\left(1-cosx\right)}{1-cos^2x}+\dfrac{1+cosx}{sinx}=\dfrac{sinx\left(1-cosx\right)}{sin^2x}+\dfrac{1+cosx}{sinx}\)

\(=\dfrac{1-cosx+1+cosx}{sinx}=\dfrac{2}{sinx}\)

\(\left(1+cotx\right)sin^3x+\left(1+tanx\right)cos^3x=sin^3x+\dfrac{cosx.sin^3x}{sinx}+cos^3x+\dfrac{sinx.cos^3x}{cosx}\)

\(=\left(sinx+cosx\right)\left(sin^2x+cos^2x-sinx.cosx\right)+sinx.cosx\left(sinx+cosx\right)\)

\(=\left(sinx+cosx\right)\left(1-sinx.cosx\right)+sinx.cosx\left(sinx+cosx\right)\)

\(=\left(sinx+cosx\right)\left(1-sinx.cosx+sinx.cosx\right)\)

\(=sinx+cosx\)