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

1.	Prove that : tan theta / sec theta +1 = sec theta -1 / tan theta
Answer» $\large\underline{\bold{Given \:Question - }}$ $\sf \: Prove \: that \: \: \dfrac{tan \theta}{sec\theta + <klux>1</klux>} = \dfrac{sec\theta - 1}{tan\theta}$ $\large\underline{\sf{Solution-}}$ Consider, $\rm :\longmapsto\:\dfrac{tan\theta}{sec\theta + 1}$ $\: \: \: \: \: \: \: \: \: \boxed{ \sf \: on \: multiply \: and \: divide \: by \: sec\theta - 1}$ $\sf \: = \: \: \: :\dfrac{tan\theta}{sec\theta + 1} \times \dfrac{sec\theta - 1}{sec\theta - 1}$ $\sf \: = \: \: \: \dfrac{tan\theta(sec\theta - 1)}{ {sec}^{2}\theta - 1 }$ $\: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf \because \: {x}^{2} - {y}^{2} = (x + y)(x - y)}$ $\sf \: = \: \: \: \dfrac{tan\theta(sec\theta - 1)}{ {tan}^{2} \theta}$ $\: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf \because \: {sec}^{2} \theta - 1 = {tan}^{2} \theta}$ $\sf \: = \: \: \: \dfrac{sec\theta - 1}{tan\theta}$ ${\boxed{\boxed{\bf{Hence, Proved}}}}$ ─━─━─━─━─━─━─━─━─━─━─━─━─ 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

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