Question

cotA/1-cotA+tanA/1-tanA= -1 prove that​

Answer 1

Step-by-step explanation:

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Answer 2

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

$\sf \: Prove \: that \: \: \dfrac{cotA}{1 - cotA} + \dfrac{tanA}{1 - tanA} \: = - 1$

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

Consider LHS,

$\sf \: \dfrac{cotA}{1 - cotA} + \dfrac{tanA}{1 - tanA}$

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

$\sf \: = \: \dfrac{\dfrac{1}{\cancel{tanA}} }{ \dfrac{tanA - 1}{\cancel{tanA}} } + \dfrac{tanA}{1 - tanA}$

$\sf \: = \: \dfrac{1}{tanA - 1} - \dfrac{tanA}{tanA - 1}$

$\sf \: = \: \dfrac{1 - tanA}{tanA - 1}$

$\sf \: = \: \dfrac{ - \: \: (\cancel{tanA - 1)}}{\cancel{tanA - 1}}$

$\sf \: = \: - \: 1$

$\large{{\boxed{\bf{Hence, Proved}}}}$

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