Question

if secA+tanA=p show that p²-1/p²+1=sinA

Answer 1

hyy friend here is ur answer
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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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#ar

Answer 2

hey here is your answer..
secA + tanA = p

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