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

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

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

prove

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

True

Solution steps

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

Manipulating left side

sec (u) cos (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{1 + sec ( x )}{1 + cos ( x )} = sec (x)

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

tan^{2} (x) = tan (2 x)

7 cos^{2} (x) + 5 sin^{2} (x) = cos (2 x) + 6

1 + sec^{2} (x) = 2 + tan^{2} (x)

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