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

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

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

prove

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

True

Solution steps

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

Manipulating left side

sec (2 a) + tan (2 a)

Express with sin, cos

sec (2 a) + tan (2 a)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply rule

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

Use the Double Angle identity:

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

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

Use the Double Angle identity:

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

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

Simplify

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

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

Apply Perfect Square Formula:

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

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

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

Apply Difference of Two Squares Formula:

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

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

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

Cancel the common factor:

sin (a) + cos (a)

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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

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

\frac{csc ^{2} ( x ) - 1}{csc ( x )} = cot (x) cos (x)

cos (- x) + sin (- x) = cos (x) - sin (x)

\frac{1 - cos ( 2 θ )}{cos ( 2 θ ) + 1} = tan^{2} (θ)

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