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Sec@ +tan@=p.,,find ,cosec@=?
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Sec@ +tan@=p.,,find ,cosec@=?
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Sec@ +tan@=p.,,find ,cosec@=?
Secθ+tanθ=p ----------------------(1)
∵, sec�θ-tan�θ=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=(p�+1)/2p
∴, cosθ=1/secθ=2p/(p�+1)
∴, sinθ=√(1-cos�θ)
=√[1-{2p/(p�+1)}�]
=√[1-4p�/(p�+1)�]
=√[{(p�+1)�-4p�}/(p�+1)�]
=√[(p⁴+2p�+1-4p�)/(p�+1)�]
=√(p⁴-2p�+1)/(p�+1)
=√(p�-1)�/(p�+1)
=(p�-1)/(p�+1)
∴, cosecθ=1/sinθ=1/[(p�-1)/(p�+1)]=(p�+1)/(p�-1) Ans.

Hope it helps you


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Sec@ +tan@=p.,,find ,cosec@=?
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