Question

10. If sec 0 + tan 0 = x, then find tan 0.​

Answer 1

Let 0 = y (for simplicity as tan 0 is like tan zero degree)

So given sec y + tan y = x

we know that $sec^2y - tan^2y = 1$

              (sec y+tany)(sec y - tan y) = 1

                 x (sec y -tan y ) = 1

                  sec y -tany = 1/x

Now we know  sec y + tan y = x

                         sec y = x -tan y

So sec y -tan y = 1/x

    x -tan y -tany = 1/x   ( substituting value of sec y)

     x -2 tan y = 1/x

 $2tan y =x - \frac{1}{x} = \frac{x^2-1}{x} \\tan y = \frac{x^2-1}{2x}$  = tan 0

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