Question

FIND. ..COSEC THETA ......

Sec THETA +tan theta =p​

Answer 1

Answer:

cosecθ = (p² + 1)/(p² - 1)

Step-by-step explanation:

Given, Secθ + tanθ = p    ------ (i)

∴ (secθ + tanθ)(secθ - tanθ) = 1

⇒ (p)(secθ - tanθ) = 1

⇒ secθ - tanθ = (1/p)     ------ (ii)

On solving (i) & (ii), we get

secθ + tanθ = p

secθ - tanθ = 1/p

-------------------------

2secθ = (p² + 1)/p

secθ = (p² + 1)/2p

∴ cosθ = (1/secθ)

           = 2p/p² + 1.

We know that sinθ = 1 - cosθ

                               = 1 - (2p)/p² + 1

                               = (p² + 1 - 2p)/(p² + 1)

                               = (p - 1)²/(p² + 1).

We know that cosecθ = 1/sinθ

                                    = (p² + 1)/(p² - 1).

Therefore, cosecθ = (p² + 1)/(p² - 1).

Hope it helps!

Answer 2

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)