Question

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

Answer 1

tan a/(1-cot a) +cot a/(1-tan a)

=(sin a/,cos a) /(1-cos a/sin a) + (cos a/sin a) /(1-sin a/cos a)

=sin ^2 a/cosa(sina - cosa) +cos^2 a/sina (cosa-sina)

=sin^2a/cosa(sina-cosa) - cos^2a/sina (sina-cosa)

=(sin^3a-cos^3a)/sina.cosa(sina-cosa)

=(sina-cosa)(sin^2a+cos^2a+sinacosa)/sina.cosa(sina-cosa)

=(1+sinacosa)/sina.cosa

=(1/sinacosa)+1

=1+seca.coseca

Answer 2

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