Is 1-cos^2(θ)=sin(θ)cos(θ)tan(θ) ?

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

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prove 1-cos^2(θ)=sin(θ)cos(θ)tan(θ)

prove

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

True

Solution steps

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

Manipulating right side

sin (θ) cos (θ) tan (θ)

Express with sin, cos

cos (θ) sin (θ) tan (θ)

Use the basic trigonometric identity:

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

= cos (θ) sin (θ) \frac{sin ( θ )}{cos ( θ )}

Simplify

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

cos (θ) sin (θ) \frac{sin ( θ )}{cos ( θ )}

Multiply fractions:

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

= \frac{sin ( θ ) cos ( θ ) sin ( θ )}{cos ( θ )}

Cancel the common factor:

cos (θ)

= sin (θ) sin (θ)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= 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

arctan (x) = \frac{1}{tan ( x )}

tan (\frac{θ}{2}) = csc (θ) - cot (θ)

cos (4 x) = 1 - 8 sin^{2} (x) + 8 sin^{4} (x)

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

\frac{sec ( x )}{csc ( x )} = tan (x)

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