Question

Prove that









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

No spamming.. solve either LHS or RHS.. DON'T solve both

Answer 1

[Tex]\huge {\pink{SOLUTION }}[/tex].

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

Answer 2

✨SOLVING L.H.S ONLY✨

$\frac{1 + { \tan }^{2} (a)}{1 + { \cot }^{2} (a)}$

$\frac{ {1}^{2} + { \tan }^{2} (a)}{ {1}^{2} + { \cot }^{2} (a)}$

$\frac{ { \sec }^{2} (a)}{ { \csc }^{2} (a)}$

$\frac{ { \sin }^{2}(a) }{ { \cos }^{2} (a)}$

$multiplying \: \: and \: \: dividing \: \: by \: \: \: \: \: \: \: \: \\ (1 -2 \sin( a) \cos(a) )$

$\frac{ (1 -2 \sin( a) \cos(a) )}{ { \cos }^{2}(a) } \times \frac{ { \sin }^{2}(a) }{ (1 -2 \sin( a) \cos(a) )}$

$( \frac{1}{ { \cos }^{2}(a) } - \frac{2 \sin(a) }{ \cos(a) } )( { \sin }^{2} (a) - \frac{ \sin(a) }{2 \cos(a) } )$

$( { \sec }^{2} (a) - 2\tan(a) )( \frac{1}{ { \csc }^{2} (a) - 2 \cot(a) }$

$\frac{{1}^{2} + { \tan }^{2} (a) - 2 \tan(a) }{{1}^{2} + { \cot }^{2} - 2 \cot(a) }$

$\frac{ ({1 - \tan(a)) }^{2} }{ {(1 - \cot(a) )}^{2} }$

$( \frac{ ({1 - \tan(a) } }{ {(1 - \cot(a) }} )^{2}$

I could've multiplied and divide when I got sec^2/csc^2 .. after solving I realised that..

it took me so much time so please:-

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