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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Apply Difference of Two Squares Formula:

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

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

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

Cancel the common factor:

sec (y) - tan (y)

= sec (y) + tan (y)

= sec (y) + tan (y)

= sec (y) + tan (y)

We showed that the two sides could take the same form

\Rightarrow True

tan (θ) (cos (θ) + cot (θ)) = sin (θ) + 1

2 cos (2 x) = 4 cos^{2} (x) - 2

tan (\frac{π}{4} - θ) = \frac{cos ( 2 θ )}{1 + sin ( 2 θ )}

\frac{1 - sin ( 2 x )}{cos ( 2 x )} = \frac{cos ( 2 x )}{1 + sin ( 2 x )}

sec (2 π) = 1 - tan^{2} (2 π)

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