Question

sinA/1+cosA+tanA/1-cosA​

Answer 1

Answer:2cosecA+tanA

Step-by-step explanation:

Answer 2

Answer:

$\frac{sinA}{1+cosA}+]\frac{tanA}{1-cosA}=2cosecA+tanA$

Step-by-step explanation:

$\frac{sinA}{1+cosA}+]\frac{tanA}{1-cosA}\\=\frac{sinA(1-cosA)}{(1+cosA)(1-cosA)}+\frac{tanA(1+cosA)}{(1-cosA)(1+cosA)}\\=\frac{sinA(1-cosA)}{1-cos^{2}A}+\frac{\frac{sinA}{cosA}(1+cosA)}{1-cos^{2}A}$

$=\frac{sinA(1-cosA)}{sin^{2}A}+\frac{sinA(1+cosA)}{cosAsin^{2}A}$

$=\frac{1-cosA}{sinA}+\frac{1+cosA}{sinAcosA}$

$=\frac{cosA(1-cosA)+1+cosA}{sinAcosA}$

$=\frac{cosA-cos^{2}A+1+cosA}{sinAcosA}$

$=\frac{cosA+(1-cos^{2}A)+cosA}{sinAcosA}$

$=\frac{2cosA+sin^{2}A}{sinAcosA}$

$=\frac{2cosA}{sinAcosA}+\frac{sin^{2}A}{sinAcosA}$

$=\frac{2}{sinA}+\frac{sinA}{cosA}\\=2cosecA+tanA$

Therefore,

$\frac{sinA}{1+cosA}+]\frac{tanA}{1-cosA}=2cosecA+tanA$

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