If sec + tan =P then find the value of Sin in terms P
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Heya !!
secA + tanA = p __(1)
We know that,
sec²A - tan²A = 1
=> (secA - tanA)(secA + tanA) = 1
Hence,
secA - tanA = 1 __(2)
Now, solving __(1) and __(2)
2secA = p+1/p
=> 2secA = (p²+1)/p
=> secA = (p²+1)/2p
Hence,
cosA = 1/secA
=> 2p/(p²+1)
We know that,
sin²A + cos²A = 1
=> sin²A = 1-cos²A
=> sinA = (1-p²)/(1+p²)
secA + tanA = p __(1)
We know that,
sec²A - tan²A = 1
=> (secA - tanA)(secA + tanA) = 1
Hence,
secA - tanA = 1 __(2)
Now, solving __(1) and __(2)
2secA = p+1/p
=> 2secA = (p²+1)/p
=> secA = (p²+1)/2p
Hence,
cosA = 1/secA
=> 2p/(p²+1)
We know that,
sin²A + cos²A = 1
=> sin²A = 1-cos²A
=> sinA = (1-p²)/(1+p²)
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