Question

Prove that : tan theta / sec theta +1 = sec theta -1 / tan theta​

Answer 1

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

$\sf \: Prove \: that \: \: \dfrac{tan \theta}{sec\theta + 1} = \dfrac{sec\theta - 1}{tan\theta}$

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

Consider,

$\rm :\longmapsto\:\dfrac{tan\theta}{sec\theta + 1}$

$\: \: \: \: \: \: \: \: \: \boxed{ \sf \: on \: multiply \: and \: divide \: by \: sec\theta - 1}$

$\sf \: = \: \: \: :\dfrac{tan\theta}{sec\theta + 1} \times \dfrac{sec\theta - 1}{sec\theta - 1}$

$\sf \: = \: \: \: \dfrac{tan\theta(sec\theta - 1)}{ {sec}^{2}\theta - 1 }$

$\: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf \because \: {x}^{2} - {y}^{2} = (x + y)(x - y)}$

$\sf \: = \: \: \: \dfrac{tan\theta(sec\theta - 1)}{ {tan}^{2} \theta}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf \because \: {sec}^{2} \theta - 1 = {tan}^{2} \theta}$

$\sf \: = \: \: \: \dfrac{sec\theta - 1}{tan\theta}$

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

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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