Question

sinA (1+tanA)+cosA (1+cotA)=(secA+cosecA)​

Answer 1

Step-by-step explanation:

sinA(1+tanA)+cosA(1+ 1/tanA)

=sinA(1+tanA)+cosA(tanA+1/tanA)

=1+tanA[sinA+cosA/tanA]

=1+tanA(sinA+cos^2A/sinA)

=[(cosA+sinA)/cosA][(sin^2A+cos^2A)/sinA]

=(cosA+sinA)/cosAsinA

=cosA/sinAcosA +sinA/sinAcosA

=1/sinA +1/cosA

=cosecA+secA

Answer 2

$sinA(1+tanA)+cosA(1+cotA)\\\\=sinA+sinAtanA+cosA+cosAtanA\\\\=sinA+sianA*\frac{sinA}{cosA}+cosA+cosA*\frac{cosA}{sinA}\\ \\= sinA+\frac{sin^{2}A}{cosA}+cosA+\frac{cos^{2}A}{sinA} \\\\=sinA+\frac{cos^{2}A}{sinA}+cosA+\frac{sin^{2}A}{cosA}\\\\=\frac{{sin^{2}A}+{cos^{2}A}}{sinA}+\frac{{cos^{2}A}+{sin^{2}A}}{cosA} \\\\=\frac{1}{sinA} +\frac{1}{cosA} \\\\=secA+cosecA$

hence proved

hope it helps!!