Question

If sec a +tan a=p ,then find the value of cosec aܿݏ݋ܿ݁ߠ

Answer 1

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

Answer:

Step-bsecθ+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.