Question

1+tanA/1-tanA=cosec^2A+2cotA/cosec^2-2

Answer 1

Answer:

Hi buddy

Step-by-step explanation:

L.H.S

1+tan^2A/1+cot^2A

=(1+sin^2A/cos2A)/(1+cos^2A/sin^2A)

=((cos^2A+sin^2A)/cos^2A)/(sin^2A+cos^2A)/sin^2A

=(1/cos^2A)/(1/sin^2A)

=1/cos^2A*sin^2A/1

=sin^2A/cos^2A

=tan^2A

M.H.S

(1-tanA/1-cotA)^2

=(1+tan^2A-2tanA)/(1+cot^2-2cotA)

=(sec^2A-2tanA)/(cosec^2A-2cosA/sinA)

=(sec^2A-2*sinA/cos A)/(cosec^2A-2cosA/sinA)

=(1/cos^2A-2sinA/cosA)/(1/sin^2A-2cosA/sinA)

=[(1-2sinAcosA)/cos^2A]/[(1-2cosAsinA)sin^2A]

=(1-2sinAcosA)/cos^2A*sin^2A/(1-2sinAcosA)

=sin^2A/cos^2A

=tan^2A

R.H.S

=tan^2A

Hence L.H.S=M.H.S=R.H.S

Hope helpful for you

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Answer 2

Ur answer is here.....!!!!!

Plz Mark as brainliest....!!!!