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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

- sin^{2} (x) + cos (x) sec (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (x)

= 1

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

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

= 1 - sin^{2} (x)

Rewrite using trig identities

1 - sin^{2} (x)

Use the Pythagorean identity:

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

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

= cos^{2} (x)

= cos^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

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

tan (θ) sec (θ) (csc (θ) - sin (θ)) = 1

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

(1 - sin^{2} (A)) (1 + tan^{2} (A)) = 1

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

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