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Prove that TanA/secA+1 +cotA/cosecA+1= cosecA+secA-secAtanA

Prove that TanA/secA+1 +cotA/cosecA+1= cosecA+secA-secAtanA

Grade:10

1 Answers

Shaswata Biswas
132 Points
7 years ago
LHS = \frac{tanA}{secA+1} + \frac{cotA}{cosecA+1}
= \frac{tanA(secA-1)}{(secA+1)(secA-1)} + \frac{cotA(cosecA-1)}{(cosecA+1)(cosecA-1)}
= \frac{tanA(secA-1)}{sec^{2}A-1} + \frac{cotA(cosecA-1)}{cosec^{2}A-1}
= \frac{tanA(secA-1)}{tan^{2}A} + \frac{cotA(cosecA-1)}{cot^{2}A}
= \frac{secA-1}{tanA} + \frac{cosecA-1}{cotA}
= \frac{secA}{tanA} - \frac{1}{tanA} + \frac{cosecA}{cotA} - \frac{1}{cotA}
= cosecA + secA - [ \frac{1}{tanA} + \frac{1}{cotA}]
= cosecA + secA - [ \frac{cosA}{sinA} + \frac{sinA}{cosA}]
= cosecA + secA - [ \frac{cos^{2}A+ sin^{2}A}{sinAcosA}]
= cosecA + secA - \frac{1}{sinAcosA}
= cosecA + secA - secAcosecA
Here it should beTanA/(secA+1) + cotA/(cosecA+1) =  cosecA +secA -secAcosecA
Hence, Proved.
THANKS
 

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