Question

(1 + cot^2 theta)/(1 + t ^ 2 * theta) = cot^2 theta​

Answer 1

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

Consider LHS

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

We know

$\red{\rm :\longmapsto\:\boxed{\tt{ tan\theta = \frac{1}{cot\theta } \: }}}$

So, on substituting the value, we get

$\rm \:  =  \: \dfrac{1 + {cot}^{2}\theta }{1 + \dfrac{1}{ {cot}^{2}\theta } }$

$\rm \:  =  \: \dfrac{1 + {cot}^{2}\theta }{\dfrac{ {cot}^{2}\theta + 1}{ {cot}^{2}\theta } }$

$\rm \:  =  \: \cancel{(1 + {cot}^{2}\theta)} \times \dfrac{ {cot}^{2}\theta }{\cancel{1 + {cot}^{2}\theta} }$

$\rm \:  =  \: {cot}^{2}\theta$

Hence,

$\rm \implies\:\boxed{\tt{ \frac{1 + {cot}^{2}\theta }{1 + {tan}^{2}\theta } = {cot}^{2}\theta \: }}$

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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