Question

cotA/1-tanA + tanA/1-cotA = 1+tanA+cosA​

Answer 1

Appropriate Question :-

Prove that

$\boxed{ \sf{ \frac{cotA}{1 - tanA} + \frac{tanA}{1 - cotA} = 1 + tanA + cotA}}$

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

Consider,

$\rm \: = \: \dfrac{cotA}{1 - tanA} + \dfrac{tanA}{1 - cotA}$

We know,

$\boxed{ \sf{cotA = \frac{1}{tanA}}}$

Using this, we get

$\rm \: = \: \dfrac{1}{tanA(1 - tanA)} + \dfrac{tanA}{1 - \dfrac{1}{tanA} }$

$\rm \: = \: \dfrac{1}{tanA(1 - tanA)} + \dfrac{tanA}{ \: \: \: \dfrac{tanA - 1}{tanA} \: \: \: }$

$\rm \: = \: \dfrac{1}{tanA(1 - tanA)} + \dfrac{tan^{2} A}{ \: \: \: tanA - 1 \: \: \: }$

$\rm \: = \: \dfrac{1}{tanA(1 - tanA)} - \dfrac{tan^{2} A}{ \: \: \: 1 - tanA \: \: \: }$

$\rm \: = \: \dfrac{1 - {tan}^{3}A }{tanA(1 - tanA)}$

We know that,

$\boxed{ \sf{ {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2}}}$

So, using this

$\rm \: = \: \dfrac{(1 - {tan}A)(1 + tanA + {tan}^{2}A)}{tanA(1 - tanA)}$

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

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

$\rm \: = \: cotA + 1 + tanA$

$\rm \: = \: 1 + tanA + cotA$

Hence,

$\boxed{ \bf{ \frac{cotA}{1 - tanA} + \frac{tanA}{1 - cotA} = 1 + tanA + cotA}}$

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

Answer 2

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