Is 2cos^2(a)=(1+sec(2a))cos(2a) ?

The answer to whether 2cos^2(a)=(1+sec(2a))cos(2a) is True

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prove 2cos^2(a)=(1+sec(2a))cos(2a)

prove

2 cos^{2} (a) = (1 + sec (2 a)) cos (2 a)

True

Solution steps

2 cos^{2} (a) = (1 + sec (2 a)) cos (2 a)

Manipulating right side

(1 + sec (2 a)) cos (2 a)

Express with sin, cos

(1 + sec (2 a)) cos (2 a)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot cos (2 a) = cos (2 a)

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (2 a)

= cos (2 a) + 1

= cos (2 a) + 1

= 1 + cos (2 a)

Rewrite using trig identities

1 + cos (2 a)

Use the Double Angle identity:

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

= 1 + 2 cos^{2} (a) - 1

Simplify

1 + 2 cos^{2} (a) - 1 : 2 cos^{2} (a)

1 + 2 cos^{2} (a) - 1

Group like terms

= 2 cos^{2} (a) + 1 - 1

1 - 1 = 0

= 2 cos^{2} (a)

= 2 cos^{2} (a)

= 2 cos^{2} (a)

We showed that the two sides could take the same form

\Rightarrow True

sec (a) - sin (a) tan (a) = cos (a)

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

sin (\frac{13 π}{2} + x) = cos (x)

tan^{2} (- θ) + 1 = sec^{2} (θ)

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

Is 2cos^2(a)=(1+sec(2a))cos(2a) ?
The answer to whether 2cos^2(a)=(1+sec(2a))cos(2a) is True