Prove that \frac{cosA-sinA+1}{cosA+sinA-1}=cosecA+cotA

byAmlesh Mahato •January 12, 2021

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Solution:

L.H.S
$=\frac{cosA-sinA+1}{cosA+sinA-1}$

⇨Dividing Numerator and Denominator By sinA

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

⇨Putting $cosec^{2}A-cot^{2}A=1$

$=\frac{cotA+cosecA-(cosec^{2}A-cot^{2}A)}{1+cotA-cosecA}$
$=\frac{cosecA+cotA-\left \{(cosecA+cotA)(cosecA-cotA) \right \}}{1+cotA-cosecA}$
$=\frac{(cosecA+cotA)+(cosecA+cotA)(cotA-cosecA)}{1+cotA-cosecA}$
$=\frac{(cosecA+cotA)(1+cotA-cosecA)}{(1+cotA-cosecA)}$
$=cosecA+cotA$
= R.H.S

Prove that \frac{cosA-sinA+1}{cosA+sinA-1}=cosecA+cotA

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