Prove that sec2θ – cos2θ = tan2θ + sin2θ - Geometry Mathematics 2

Question

Prove that sec²θ – cos²θ = tan²θ + sin²θ

Sum

Solution

L.H.S = sec²θ – cos²θ

= sec²θ – (1 – sin²θ) ...... $[\begin{matrix} ∵ \sin^{2} θ + \cos^{2} θ = 1 \\ ∴ 1 - \sin^{2} θ = \cos^{2} θ \end{matrix}]$

= sec²θ – 1 + sin²θ

= tan²θ + sin²θ ...... $[\begin{matrix} ∵ 1 + \tan^{2} θ = \sec^{2} θ \\ ∴ \tan^{2} θ = \sec^{2} θ - 1 \end{matrix}]$

= R.H.S

∴ sec²θ – cos²θ = tan²θ + sin²θ

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Prove that sec2θ – cos2θ = tan2θ + sin2θ - Geometry Mathematics 2

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