Question

sec theata +tan theata=p. find cosect theatha

Answer 1

Step-by-step explanation:

secθ+tanθ=p ----------------------(1)

∵, sec²θ-tan²θ=1

or, (secθ+tanθ)(secθ-tanθ)=1

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) Ans.

Answer 2

sec + tan = p

1/cos + sin /cos = p

(sin +1)/cos = p

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

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

sin + 1 = p^2 - p^2sin

sin + p^2 sin = p^2 -1

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

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

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