Is sin(x)cos(x)tan(x)=1-cos^2(x) ?

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

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prove sin(x)cos(x)tan(x)=1-cos^2(x)

prove

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

True

Solution steps

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

Manipulating left side

sin (x) cos (x) tan (x)

Express with sin, cos

cos (x) sin (x) tan (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (x)

= sin (x) sin (x)

Apply exponent rule:

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

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

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

= sin^{2} (x)

= sin^{2} (x)

Rewrite using trig identities

sin^{2} (x)

Use the Pythagorean identity:

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

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

= 1 - cos^{2} (x)

= 1 - cos^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

\frac{1 + tan ( u )}{1 + cot ( u )} = tan (u)

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

3 sin^{2} (2 x) = 12 cos^{2} (x) sin^{2} (x)

cos (θ) = sin (θ) cot (θ)

sin (θ) cos (θ) = 1

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