Is sin(θ)*tan(θ)*cos(θ)=sin^2(θ) ?

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

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prove sin(θ)tan(θ)cos(θ)=sin^2(θ)

prove

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

True

Solution steps

sin (θ) tan (θ) cos (θ) = sin^{2} (θ)

Manipulating left side

sin (θ) tan (θ) cos (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

p ro v e - sin (θ) = - sin (- θ)

p ro v e cos (2 B) = \frac{1 - tan ^{2} ( B )}{1 + tan ^{2} ( B )}

p ro v e \frac{1 + sec ( x )}{1 + cos ( x )} = sec (x)

p ro v e \frac{1 - sin ^{2} ( x )}{sin ( x )} = cot (x) cos (x)

p ro v e 7 cos^{2} (x) + 5 sin^{2} (x) = cos (2 x) + 6

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