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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