Question

tan A/(1-cotA)+cot A/(1-tanA)=1+secA×cosecA​

Answer 1

Answer:

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

Step-by-step explanation:

hope it helps you ❣️

Answer 2

Answer

$\rm \dfrac{tan\ A}{1-cot\ A}+\dfrac{cot\ A}{1-tan\ A}=1+sec\ A.cosec\ A$

Given

$\rm \dfrac{tan\ A}{1-cot\ A}+\dfrac{cot\ A}{1-tan\ A}$

To Prove

$\rm 1+sec\ A.cosec\ A$

Solution

$\rm LHS\\\\\to \rm \dfrac{tan\ A}{1-cot\ A}+\dfrac{cot\ A}{1-tan\ A}\\\\\to \rm \dfrac{tan\ A}{1-\frac{1}{tan\ A}}+\dfrac{cot\ A}{1-tan\ A}\\\\\to \rm \dfrac{tan\ A\times tan\ A}{tan\ A-1}+\dfrac{cot\ A}{1-tan\ A}\\\\\to \rm \dfrac{tan^2A}{tan\ A-1}+\dfrac{cot\ A}{1-tan\ A}\\\\\to \rm \dfrac{cot\ A}{1-tan\ A}-\dfrac{tan^2A}{1-tan\ A}\\\\\to \rm \dfrac{\frac{1}{tan\ A}}{1-tan\ A}-\dfrac{tan^2A}{1-tan\ A}\\\\\to \rm \dfrac{1-tan^3A}{tan\ A(1-tan\ A)}$

$\to \rm \dfrac{\cancel{(1-tan\ A)}(1+tan\ A+tan^2A)}{tan\ A\cancel{(1-tan\ A)}}\\\\\to \rm \dfrac{1+tan\ A+tan^2A}{tan\ A}\\\\\to \rm \dfrac{tan\ A}{tan\ A}+\dfrac{1+tan^2A}{tan\ A}\\\\\to \rm 1+\dfrac{sec^2A}{tan\ A}\ [\; \because sec^2A-tan^2A=1 \;]\\\\\to \rm 1+sec\ A.sec\ A.\dfrac{cos\ A}{sin\ A}\\\\\to \rm 1+sec\ A.csc\ A\\\\\to \rm RHS$