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

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

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

prove

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

True

Solution steps

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

Manipulating right side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

- cos^{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

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

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

= 1 - cos^{2} (x)

Rewrite using trig identities

1 - cos^{2} (x)

Use the Pythagorean identity:

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

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

= sin^{2} (x)

= sin^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

p ro v e 1 - 2 sin^{2} (2 θ) = 8 cos^{4} (θ) - 8 cos^{2} (θ) + 1

p ro v e sech^{2} (x) + tanh^{2} (x) = 1

p ro v e 2 cos^{2} (A) - cos (2 A) - 1 = 0

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

p ro v e csc (A) = \frac{7}{4}

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