What is the general solution for 1/((sec(a)-tan(a)))=sec(a)+tan(x) ?

The general solution for 1/((sec(a)-tan(a)))=sec(a)+tan(x) is x=arctan(1/(sec(a)-tan(a))-sec(a))+pin

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1/((sec(a)-tan(a)))=sec(a)+tan(x)

\frac{1}{( sec ( a ) - tan ( a ) )} = sec (a) + tan (x)

x = arctan (\frac{1}{sec ( a ) - tan ( a )} - sec (a)) + πn

Solution steps

\frac{1}{( sec ( a ) - tan ( a ) )} = sec (a) + tan (x)

Switch sides

sec (a) + tan (x) = \frac{1}{sec ( a ) - tan ( a )}

Move

sec (a)

to the right side

tan (x) = \frac{1}{sec ( a ) - tan ( a )} - sec (a)

Apply trig inverse properties

x = arctan (\frac{1}{sec ( a ) - tan ( a )} - sec (a)) + πn

- 2 = tan^{2} (x)

3 sin^{2} (x) - 1 = cos^{4} (x)

sin (3 \cdot x) = cos (x)

sin^{22} (x) = sin^{2} (x)

sin^{2} (2 x) + cos^{2} (x) - 1 = 0

What is the general solution for 1/((sec(a)-tan(a)))=sec(a)+tan(x) ?
The general solution for 1/((sec(a)-tan(a)))=sec(a)+tan(x) is x=arctan(1/(sec(a)-tan(a))-sec(a))+pin