CLASS X
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Please find the answer in the attachment.
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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) & (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
= √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)]
cosecθ =(p²+1)/(p²−1)
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