Question

1+tan²A/1+cot²A=(1-tanA/1-cotA)²​

Answer 1

$\large\underline{\bold{Given \:Question - }}$

$\sf \: Prove \: that \: \dfrac{1 + {tan}^{2}A }{1 + {cot}^{2}A } = {\bigg(\dfrac{1 - tanA}{1 - cotA} \bigg) }^{2}$

$\large\underline{\sf{Answer-}}$

Consider,

$\rm :\longmapsto\:\dfrac{1 + {tan}^{2}A }{1 + {cot}^{2}A }$

$\sf \: = \: \dfrac{1 + {tan}^{2}A }{1 + \dfrac{1}{ {tan}^{2}A }}\: \: \: \: \{ \because \: cotA = \dfrac{1}{tanA} \}$

$\sf \: = \: \dfrac{1 + {tan}^{2}A }{\dfrac{ {tan}^{2}A + 1 }{ {tan}^{2}A } }$

$\sf \: = \: \cancel{({1 + tan}^{2}A)} \times \dfrac{ {tan}^{2}A }{\cancel{(1 + {tan}^{2}A} }$

$\sf \: = \: {tan}^{2} A \: - - - - (1)$

Consider

$\rm :\longmapsto\:{\bigg(\dfrac{1 - tanA}{1 - cotA} \bigg) }^{2}$

$\sf \: = \: {\bigg(\dfrac{1 - tanA}{1 - \dfrac{1}{tanA}} \bigg) }^{2} \: \: \: \: \{ \because \: cotA = \dfrac{1}{tanA} \}$

$\sf \: = \: {\bigg(\dfrac{1 - tanA}{\dfrac{tanA - 1}{tanA}} \bigg) }^{2}$

$\sf \: = \: {\bigg( - \cancel{(tanA - 1)} \times \dfrac{tanA}{\cancel{tanA - 1}} \bigg) }^{2}$

$\sf \: = \: {\bigg( - tanA\bigg) }^{2}$

$\sf \: = \: {tan}^{2} A - - - - (2)$

From (1) and (2), we concluded that

$\rm :\longmapsto\:\sf \: \: \dfrac{1 + {tan}^{2}A }{1 + {cot}^{2}A } = {\bigg(\dfrac{1 - tanA}{1 - cotA} \bigg) }^{2}$

Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1