Question

sec A + tan A = X then tan A = ?

Answer 1

secA + tanA = x ... ( 1 ) 
=> 1/(secA - tanA) = x as (sec A + tan A)(sec A - tan A) = sec^2 A - tan ^2 A = 1 
=> secA - tanA = 1/x ... ( 2) 
Subtracting Eqn. ( 2 ) from eqn. ( 1 ), 
2tanA = x - 1/x 
=> tanA = (1/2) (x - 1/x). 

Answer 2

Answer: secA + tanA = X........(1)

We know that :

1 + tanA^2 = secA^2

1 = sec^2A - tan^2A

1 = (secA + tanA) (secA - tanA)

secA + tanA = 1/secA - tanA

So,

X = 1/secA - tanA

SecA - tanA = 1/X..........(2)

Subtract eq^n2 from eq^n1

secA + tanA = X

secA - tanA = 1/X

- + -

tanA = x - 1/x

tanA = (x^2 - 1)/x

Step-by-step explanation: