Is 1-cos(θ)sin(θ)tan(θ)=cos^2(θ) ?

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

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prove 1-cos(θ)sin(θ)tan(θ)=cos^2(θ)

prove

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

True

Solution steps

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

Manipulating left side

1 - cos (θ) sin (θ) tan (θ)

Express with sin, cos

1 - cos (θ) sin (θ) tan (θ)

Use the basic trigonometric identity:

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

= 1 - cos (θ) sin (θ) \frac{sin ( θ )}{cos ( θ )}

Simplify

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

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

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)

= 1 - sin^{2} (θ)

= 1 - sin^{2} (θ)

= 1 - sin^{2} (θ)

Rewrite using trig identities

1 - sin^{2} (θ)

Use the Pythagorean identity:

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

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

= cos^{2} (θ)

= cos^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

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

\frac{sin ( 6 0 ^{\circ} )}{cos ( 6 0 ^{\circ} )} = tan (6 0^{\circ})

cos (x) csc (x) = cos (x)

cot (x) + \frac{1}{cot ( x )} = \frac{2}{sin ( 2 x )}

sec^{2} (t) = cos^{- 2} (t)

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