If secA + tanA = p then prove that p^2-1 by p^2+1 = sin A
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SecA+tanA=p ----------------------------(1)
We know that,
sec²A-tan²A=1
or, (secA+tanA)(secA-tanA)=1
or, p(secA-tanA)=1
or, secA-tanA=1/p -----------------------(2)
Adding (1) and (2) we get,
2secA=p+1/p
or, 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) (Proved)
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