Question

If secA+tanA=p, then prove that sinA=
$\frac{ {p }^{2} - 1 }{ {p}^{2} + 1 }$



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Answer 1

Step-by-step explanation:

secA + tanA = p

squaring both sides

(secA + tanA)^2 = p^2

sec^2A + tan^2A + 2secA.tanA = p^2

1/cos^2A + sin^2A/cos^2A + 2sinA/cos^2A = p^2

Take cos^2A LCM

(1 + sin^2A + 2 sinA) / cos^2A = p^2

(cos^2A = 1 - sin^2A)

(1 + sinA)^2 / (1 - sin^2A) = p^2

(1 + sinA)^2 / (1 + sinA)(1 - sinA) = p^2

(1 + sinA) / (1 - sinA) = p^2

by cross multiplication

1 + sinA = p^2 - p^2.sinA

sinA + p^2.sinA = (p^2 - 1)

sinA(1 + p^2) = (p^2 - 1)

sinA = (p^2 - 1)/(1 + p^2)