IF secthita+ tanthita =p Then find …
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Posted by Tamman Patnaik 6 years, 11 months ago
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Gaurav Seth 6 years, 11 months ago
Sec2θ - tan2θ = 1
(secθ+tanθ)(secθ-tanθ)=1
<hr />or, secθ-tanθ=1/p ----------------(2)
Adding (1) and (2) we get,
2secθ=p+1/p
or, secθ=(p²+1)/2p
∴, cosθ=1/secθ=2p/(p²+1)
∴, sinθ=√(1-cos²θ)
=√[1-{2p/(p²+1)}²]
=√[1-4p²/(p²+1)²]
=√[{(p²+1)²-4p²}/(p²+1)²]
=√[(p⁴+2p²+1-4p²)/(p²+1)²]
=√(p⁴-2p²+1)/(p²+1)
=√(p²-1)²/(p²+1)
=(p²-1)/(p²+1)
∴, cosecθ=1/sinθ=1/[(p²-1)/(p²+1)]=(p²+1)/(p²-1)
Another method:
Sec2θ - tan2θ = 1
(secθ + tanθ)( secθ - tanθ) = 1
p( secθ - tanθ) = 1
( secθ - tanθ) = 1/p and, ......-(1)
( secθ + tanθ) = p ..... - (2)
(1) + (2),
secθ = 1/2 (p + 1/p) ,
(2) - (1)
tanθ = 1/2 (p-1/p)
secθ / tanθ = cosecθ
Hence,
cosecθ = p +1/p / p-1/p = p2 +1 / p2 -1
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