Question

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Please help me with this question.

If tan theta + sec theta =p
Then find the value of cosec theta.

Answer 1

Answer:

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.

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Answer 2

$\huge \mathtt{heya }$

here is ur answer ➖➖➖➖➖➖➖➖⬇

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.


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