Is sec(x)+tan(x)= 1/(sec(x)-tan(x)) ?

The answer to whether sec(x)+tan(x)= 1/(sec(x)-tan(x)) is True

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

prove

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

True

Solution steps

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

Manipulating right side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

tan^{2} (x) + 1 = sec^{2} (x)

1 = sec^{2} (x) - tan^{2} (x)

= \frac{sec ^{2} ( x ) - tan ^{2} ( x )}{sec ( x ) - tan ( x )}

Simplify

\frac{sec ^{2} ( x ) - tan ^{2} ( x )}{sec ( x ) - tan ( x )} : sec (x) + tan (x)

\frac{sec ^{2} ( x ) - tan ^{2} ( x )}{sec ( x ) - tan ( x )}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

sec^{2} (x) - tan^{2} (x) = (sec (x) + tan (x)) (sec (x) - tan (x))

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

Cancel the common factor:

sec (x) - tan (x)

= sec (x) + tan (x)

= sec (x) + tan (x)

= sec (x) + tan (x)

We showed that the two sides could take the same form

\Rightarrow True

\frac{sec ^{2} ( x ) - 1}{sec ^{2} ( x )} = sin^{2} (x)

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

(sec^{2} (x) - 1) csc (x) = tan (x) sec (x)

cos (x - \frac{π}{2}) = sin (x)

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

Is sec(x)+tan(x)= 1/(sec(x)-tan(x)) ?
The answer to whether sec(x)+tan(x)= 1/(sec(x)-tan(x)) is True