If secx+tanx=p, then which of the following is/are correct?
A
sinx=p2−1p2+1
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B
cosx=2pp2+1
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C
tanx=p2−12p
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D
secx=p2+12p
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Solution
The correct option is Dsecx=p2+12p secx+tanx=p⋯(1)
We know that, sec2x−tan2x=1⇒secx−tanx=1secx+tanx⇒secx−tanx=1p⋯(2)
From equation (1) and (2), 2secx=p+1p⇒secx=p2+12p⇒cosx=2pp2+1
And 2tanx=p−1p⇒tanx=p2−12p
Now, sinx=tanx×cosx=p2−1p2+1