Math, asked by sauryashivam, 10 months ago

If sec 0 + tan 0 =p, prove that
sin 0 = p^2 -1/p^2+1​

Answers

Answered by Amy246
1

Step-by-step explanation:

sec + tan = p

Sec = p - tan

sec^2 = (p - tan)^2

sec^2 = p^2 + tan^2 - 2ptan

sec^2 - tan^2 = p^2 - 2ptan

p^2 - 2ptan - 1 = 0

(p^2 - 1)/2p = tan

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

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

Cos = 2p/(p^2+1)

cos^2 = 4p^2/(p^2+1)^2

sin^2 = 1 - 4p^2/(p^2+1)^2

sin^2 =[ (p^2+1)^2 - 4p^2]/(p^2+1)^2

sin^2 = (p^2–1)^2/(p^2+1)^2

Sin = (p^2–1)/(p^2+1)

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