If sec -tan-p- prove that sin- p 2-1 p 2-1

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Value of Standard Angle(30,60,90)

If sec θ+tanθ...

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If sec θ + tan θ = p, prove that sin θ = $\frac{p^{2} - 1}{p^{2} + 1}$

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s e c θ + \tan θ = p, show that, \frac{p^{2} - 1}{p^{2} + 1} = \sin θ

\sec θ + \tan θ = p

(i) \sec θ = \frac{1}{2} (p + \frac{1}{p}) (ii) \tan θ = \frac{1}{2} (p - \frac{1}{p}) (iii) \sin θ = \frac{p^{2} - 1}{p^{2} + 1}

If secθ+tanθ=p ,prove that = (p²-1)/(p²+1).

\frac{1}{p}

\frac{2 p}{(p^{2} + 1)} and \sin θ = \frac{p^{2} - 1}{p^{2} + 1}

(sec θ + tan θ) = p

sin θ = \frac{p^{2} - 1}{p^{2} + 1}

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