Is cos(u)sec(u)-cos^2(u)=sin^2(u) ?

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

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prove cos(u)sec(u)-cos^2(u)=sin^2(u)

prove

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

True

Solution steps

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

Manipulating left side

cos (u) sec (u) - cos^{2} (u)

Express with sin, cos

- cos^{2} (u) + cos (u) sec (u)

Use the basic trigonometric identity:

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

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

Simplify

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

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

cos (u) \frac{1}{cos ( u )} = 1

cos (u) \frac{1}{cos ( u )}

Multiply fractions:

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

= \frac{1 \cdot cos ( u )}{cos ( u )}

Cancel the common factor:

cos (u)

= 1

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

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

= 1 - cos^{2} (u)

Rewrite using trig identities

1 - cos^{2} (u)

Use the Pythagorean identity:

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

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

= sin^{2} (u)

= sin^{2} (u)

We showed that the two sides could take the same form

\Rightarrow True

\frac{sec ( θ )}{cot ( θ ) + tan ( θ )} = sin (θ)

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

\frac{cos ( θ ) + 1}{sin ( θ )} = cot (θ) + csc (θ)

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

2 sin^{2} (3 x) = 1 - cos (6 x)

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