Question

if secO+tanO=p , then prove that sinO = p2-1/p2+1​

Answer 1

Answer:

secA+tanA=p -------(1)

We know that,

sec²A - tan²A = 1

(secA + tanA )(secA - tanA) = 1

p(secA - tanA) = 1

secA - tanA = 1/p ------------(2)

Adding (1) and (2) we get,

2secA = p+1/p

secA = (p²+1)/2p

cosA = 1/secA = 2p/(p²+1)

sinA = √(1-cos²A)

=√{1-4p²/(p²+1)²

=√{(p²+1)²-4p²}/(p²+1)²

=√(p⁴+2p²+1-4p²)/(p²+1)

=√(p²-1)²/(p²+1)

=(p²-1)/(p²+1)

Hence proved

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