Is sec(2x)+tan(2x)=((cos(x)+sin(x)))/((cos(x)-sin(x))) ?

The answer to whether sec(2x)+tan(2x)=((cos(x)+sin(x)))/((cos(x)-sin(x))) is True

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prove sec(2x)+tan(2x)=((cos(x)+sin(x)))/((cos(x)-sin(x)))

prove

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

True

Solution steps

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

Manipulating left side

sec (2 x) + tan (2 x)

Express with sin, cos

sec (2 x) + tan (2 x)

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

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

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

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

Simplify

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

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

Apply rule

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

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

Use the Double Angle identity:

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

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

Simplify

\frac{cos ^{2} ( x ) + sin ^{2} ( x ) + 2 sin ( x ) cos ( x )}{cos ^{2} ( x ) - sin ^{2} ( x )} : \frac{sin ( x ) + cos ( x )}{cos ( x ) - sin ( x )}

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

Apply Perfect Square Formula:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = sin (x), b = cos (x)

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

Apply Difference of Two Squares Formula:

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

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

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

Cancel the common factor:

sin (x) + cos (x)

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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

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

\frac{1 + sec ( x )}{csc ( x )} = cot (x) cos (x)

cos ((\frac{π}{2} - x) - 4) = sin (x + y)

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

Is sec(2x)+tan(2x)=((cos(x)+sin(x)))/((cos(x)-sin(x))) ?
The answer to whether sec(2x)+tan(2x)=((cos(x)+sin(x)))/((cos(x)-sin(x))) is True