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Differentiation Using Substitution

Prove that - ...

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Prove that - tan + tan = sec + sec

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{tan}^{2} A {sec}^{2} B - {sec}^{2} A {tan}^{2} B = {tan}^{2} A - {tan}^{2} B

\frac{tan θ + sec θ - 1}{tan θ - sec θ - 1} = sec θ + tan θ

Q. If (sec θ – tan θ) =

\sqrt{2}

tan θ then prove that (sec θ + tan θ) =

\sqrt{2}

(sec θ + tan θ - 1) (sec θ - tan θ + 1) = 2 tan θ

(sec A + tan A - 1) (sec A - tan A + 1) = 2 tan A

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