prove that (1+tan^2aupon 1+cot^2a)=(1-tan^2a upon 1-cot^2 a)=tan^2a
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Answer:
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
Step-by-step explanation:
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