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If sec A+tanA=x , the secA=?
Dear Nitin,
Given that SecA+TanA=x------ Eq.1
We know that Sec2A-Tan2A=1
This is of the form a2-b2=(a+b)(a-b)
So (SecA-TanA)(SecA+TanA)=1
SecA-TanA=1/x {From eq.1}
Lets consider this as Eq.2
Now adding Eq.1 and Eq.2, we get 2SecA=x+1/x
Hence SecA=(x+1/x)/2
So SecA=(x2+1)/2x
All the best!
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