Question

sin/1-cos+tan/1+cos=sec.cosec+cot

Answer 1

Answer:

Step-by-step explanation:

Very simple answer ,just needed a little understanding .

Answer 2

Answer:

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

Step-by-step explanation:

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

$=\frac{sinA+sinAcosA+\frac{sinA}{cosA}+\frac{sinA}{cosA}\times cosA}{sin^{2}A}\\=\frac{sinA+sinAcosA+\frac{sinA}{cosA}-sinA}{sin^{2}A}\\=\frac{sinAcosA+\frac{sinA}{cosA}}{sin^{2}A}\\=\frac{sinAcosA}{sin^{2}A}+\frac{sinA}{sin^{2}AcosA}\\=\frac{cosA}{sinA}+\frac{1}{sinAcosA}\\=cotA+cosecAsecA\\=RHS$

Therefore,.

$\frac{sinA}{1-cosA}+\frac{tanA}{1+cosA}\\=cotA+cosecA secA$

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