Question

(1+tanA/1-cotA)square=tansquareA

Answer 1

$\boxed{\huge{\purple{Answer:-}}}$

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$\boxed{\bf{L.H.S}}$

(1-tanA/1-cotA)²

=(1+tan²A-2tanA)/(1+cot²-2cotA)

=(sec²A-2tanA)/(cosec²A-2cosA/sinA)

=(sec²A-2*sinA/cos A)/(cosec²A-2cosA/sinA)

=(1/cos²A-2sinA/cosA)/(1/sin²A-2cosA/sinA)

=[(1-2sinAcosA)/cos²A]/[(1-2cosAsinA)sin²A]

=(1-2sinAcosA)/cos²A×sin²A/(1-2sinAcosA)

=sinA/cos²A

=tan²A

$\boxed{\bf{R.H.S}}$

=tan²A

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Hope it help you ✌ ☺