Is sin^2(θ)cos(θ)sec(θ)=1-cos^2(θ) ?

The answer to whether sin^2(θ)cos(θ)sec(θ)=1-cos^2(θ) is True

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prove sin^2(θ)cos(θ)sec(θ)=1-cos^2(θ)

prove

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

True

Solution steps

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

Manipulating left side

sin^{2} (θ) cos (θ) sec (θ)

Express with sin, cos

cos (θ) sec (θ) sin^{2} (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (θ)

= 1 \cdot sin^{2} (θ)

Multiply:

1 \cdot sin^{2} (θ) = sin^{2} (θ)

= sin^{2} (θ)

= sin^{2} (θ)

= sin^{2} (θ)

Rewrite using trig identities

sin^{2} (θ)

Use the Pythagorean identity:

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

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

= 1 - cos^{2} (θ)

= 1 - cos^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

sin (x) = \frac{cos ( x )}{cot ( x )}

tan^{2} (θ) - tan^{2} (θ) sin^{2} (θ) = sin^{2} (θ)

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

(1 + sin (z)) (1 - sin (z)) = \frac{1}{sec ^{2} ( z )}

\frac{tan ( x )}{sec ( x ) + 1} = csc (x) - cot (x)

Is sin^2(θ)cos(θ)sec(θ)=1-cos^2(θ) ?
The answer to whether sin^2(θ)cos(θ)sec(θ)=1-cos^2(θ) is True