Question

tanA/1-cotA+cotA/1-tanA=1+secA cosecA​

Answer 1

$\frac{tanA}{1 - tanA} + \frac{cotA}{1 - tanA} \\ \\ = \frac{ \frac{cosA}{sinA} }{ \frac{cosA - sinA}{cosA} } + \frac{ \frac{sinA}{cosA} }{ \frac{sinA - cosA}{sinA} } \\ \\ = \frac{ {cos}^{2}A }{sinA(cosA - sinA)} + \frac{ {sin}^{2}A }{cosA(sinA - cosA)} \\ \\ = \frac{ {cos}^{2} A}{sinA(cosA - sinA)} - \frac{ {sin}^{2}A }{cosA(cosA - sinA)} \\ \\ = \frac{ {cos}^{3}A- {sin}^{3} A }{sinAcosA(cosA - sinA)} \\ \\ = \frac{(cosA - sinA)(1 + sinAcosA)}{sinAcosA(cosA - sinA)} \\ \\ = \frac{1 + sinAcosA}{sinAcosA} \\ \\ = 1 + secAcosecA$