Question

sinA/1+cotA - cosA/1+TanA=SinA-CosA​

Answer 1

Answer:

LHS:

sin A/ 1+cot A - cosA/1+tanA

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

$sinA$ × $\frac{sinA}{sinA+cosA} -$ $cos A$ × $\frac{cos A}{sin A + cosA}$

$\frac{sin^{2}A - cos^{2}A }{sinA + cosA}$

$\frac{(sinA+cosA)(sinA-cos)}{sinA+cosA} \\\\sinA-cosA$

$= RHS$ (proved)

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Answer 2

$\frac{sinA}{1 + cotA} - \frac{cosA}{1 + tanA} \\ \\ = \frac{ {sin}^{2} A}{sinA + cosA} - \frac{ {cos}^{2}A }{cosA + sinA} \\ \\ = \frac{ {sin}^{2}A - {cos}^{2} A }{sinA + cosA} \\ \\ = sinA - cosA$