Question

sinA/1-cosA+tanA/1+cosA=secA×cosecA+cotA

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

$\frac{sinA}{1-cosA}+\frac{tan}{1+cosA}=cotA+cosecAsecA$

Step-by-step explanation:

$LHS=\frac{sinA}{1-cosA}+\frac{tan}{1+cosA}\\=\frac{sinA(1+cosA)}{(1-cosA)(1+cosA)}+\frac{tanA(1-cosA)}{(1+cosA)(1-cosA)}\\=\frac{sinA(1+cosA)}{sin^{2}A}+\frac{\frac{sinA}{cosA}\times(1-cosA)}{sin^{2}A}\\=\frac{1+cosA}{sinA}+\frac{1-cosA}{cosAsinA}\\=\frac{1}{sinA}+\frac{cosA}{sinA}+\frac{1}{sinAcosA}-\frac{cosA}{sinAcosA}\\=\frac{1}{sinA}+cotA+\frac{1}{sinAcosA}-\frac{1}{sinA}\\=cotA+\frac{1}{sinA}\cdot \frac{1}{cosA}\\=cotA+cosecAsecA\\=RHS$

Therefore,

$\frac{sinA}{1-cosA}+\frac{tan}{1+cosA}=cotA+cosecAsecA$

•••♪