Step-by-step explanation:
[tex]\frac{sinA}{1+cosA} +\frac{1+cos}{sinA} =2cosec\\\\\\\frac{sin^{2}A+(1+cos)^{2} }{sinA(1+cosA)} \\\\[/tex]
[tex]\frac{sin^{2}A+1+cos^{2}+2cos}{sinA(1+cosA)} \\\\[/tex] and we have [tex]sin^{2}+cos^{2} =1[/tex]
so [tex]\frac{1+1+2cosA }{sinA(1+cosA)} \\\\[/tex]
[tex]\frac{2+2cosA }{sinA(1+cosA)} \\\\[/tex]
[tex]\frac{2(1+cosA)}{sinA(1+cosA)} \\\\[/tex]
[tex]\frac{2}{sinA}[/tex] we have [tex]\frac{1}{sin}=cosec\\[/tex]
Finally [tex]2cosecA[/tex]
I really hope this helps coz it took much time
