To Prove: (1+cotA + tanA)(sinA - cosA) = \(\frac{secA}{cosec^2A}-\frac{cosecA}{sec^2A}\) = sinAtanA - cotAcosA
Proof:Consider the LHS,
= \(\frac{secA}{cosec^2A}-\frac{cosecA}{sec^2A}\)
= \(\Big(\frac{1}{cosA}\times sin^2A\Big)-\Big(\frac{1}{sinA}\times cos^2A\Big)\)
= sinA - cosA+ cotA sinA - cotA cosA + tanA sinA - tanA cosA
Use the formula:
tanθ=\(\frac{sinθ}{cosθ}\) and cotθ = \(\frac{cosθ}{sinθ}\)
= sinA - cosA + \(\Big(\frac{cosA}{sinA}\times sinA\Big)\)-\(\Big(\frac{cosA}{sinA}\times cosA\Big)\) + \(\Big(\frac{sinA}{cosA}\times sinA\Big)\)-\(\Big(\frac{sinA}{cosA}\times cosA\Big)\)
= sinA - cosA + cosA - \(\frac{cos^2A}{sinA}+\frac{sin^2A}{cosA}\) - sinA
= \(\frac{sin^2A}{cosA}-\frac{cos^2A}{sinA}\)
We know:
Again use the formula:
Use the formula:
Hence Proved.
