Question

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

Answer 1

Answer:

$\frac{\textstyle 1+\tan A}{\textstyle 1-\tan A}\\\\=\frac{\textstyle \cos A(1+\tan A)}{\textstyle\cos A(1-\tan A)}\\\\= \frac{\textstyle \cos A + \sin A}{\textstyle\cos A - \sin A}\\\\=\frac{\textstyle(\cos A + \sin A)(\cos A+\sin A)}{\textstyle(\cos A - \sin A)(\cos A+\sin A)}\\\\=\frac{\textstyle\cos^2A+\sin^2A+2\cos A\sin A}{\textstyle\cos^2A-\sin^2A}$

$=\frac{\textstyle1+2\cos A\sin A}{\textstyle(1-\sin^2A)-\sin^2A}\\\\=\frac{\textstyle1+2\cos A\sin A}{\textstyle1-2\sin^2A}\\\\=\frac{\textstyle\sin^2A(\mathrm{cosec}^2A+2\cot A)}{\textstyle\sin^2A(\mathrm{cosec}^2A-2)}\\\\=\frac{\textstyle\mathrm{cosec}^2A+2\cot A}{\textstyle\mathrm{cosec}^2A-2}$

Hope that helps!

Answer 2

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

r.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= l.h.s

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