Is cos(θ)(sec(θ)-cos(θ))=sin^2(θ) ?

The answer to whether cos(θ)(sec(θ)-cos(θ))=sin^2(θ) is True

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prove cos(θ)(sec(θ)-cos(θ))=sin^2(θ)

prove

cos (θ) (sec (θ) - cos (θ)) = sin^{2} (θ)

True

Solution steps

cos (θ) (sec (θ) - cos (θ)) = sin^{2} (θ)

Manipulating left side

cos (θ) (sec (θ) - cos (θ))

Express with sin, cos

(- cos (θ) + sec (θ)) cos (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

(- cos (θ) + \frac{1}{cos ( θ )}) cos (θ)

Join

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

- cos (θ) + \frac{1}{cos ( θ )}

Convert element to fraction:

cos (θ) = \frac{cos ( θ ) cos ( θ )}{cos ( θ )}

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

Since the denominators are equal, combine the fractions:

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

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

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

- cos (θ) cos (θ) + 1

cos (θ) cos (θ) = cos^{2} (θ)

cos (θ) cos (θ)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= cos^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= cos^{2} (θ)

= - cos^{2} (θ) + 1

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (θ)

= - - cos^{2} (θ) + 1

= - cos^{2} (θ) + 1

= 1 - cos^{2} (θ)

Rewrite using trig identities

1 - cos^{2} (θ)

Use the Pythagorean identity:

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

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

= sin^{2} (θ)

= sin^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

p ro v e \frac{2 cot ( x )}{csc ^{2} ( x )} = sin (2 x)

p ro v e cos^{4} (x) - sin^{4} (x) = 2 cos^{2} (x) - 1

p ro v e tan (x) (cot (x) + tan (x)) = sec^{2} (x)

p ro v e \frac{1 + sin ( 2 x )}{sin ( 2 x )} = 1 + \frac{1}{2} sec (x) csc (x)

p ro v e \frac{1}{csc ( x ) + 1} + \frac{1}{sin ( x ) + 1} = 1

Is cos(θ)(sec(θ)-cos(θ))=sin^2(θ) ?
The answer to whether cos(θ)(sec(θ)-cos(θ))=sin^2(θ) is True