Is cos(θ)tan(θ)sin(θ)=sin^2(θ) ?

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

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prove cos(θ)tan(θ)sin(θ)=sin^2(θ)

prove

cos (θ) tan (θ) sin (θ) = sin^{2} (θ)

True

Solution steps

cos (θ) tan (θ) sin (θ) = sin^{2} (θ)

Manipulating left side

cos (θ) tan (θ) sin (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)

We showed that the two sides could take the same form

\Rightarrow True

sec (x) tan (x) (1 - sin^{2} (x)) = sin (x)

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

cos (\frac{3 x}{5}) = - sin (\frac{3 x}{5} - \frac{π}{2})

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

\frac{( cos ( a - b ) )}{sin ( a ) sin ( b )} = cot (a) cot (b) + 1

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