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

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

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

prove

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

True

Solution steps

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

Manipulating left side

1 - sin (x) cos (x) tan (x)

Express with sin, cos

1 - cos (x) sin (x) tan (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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)

= 1 - sin^{2} (x)

= 1 - sin^{2} (x)

= 1 - sin^{2} (x)

Rewrite using trig identities

1 - sin^{2} (x)

Use the Pythagorean identity:

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

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

= cos^{2} (x)

= cos^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

sin (x) cos (x) = sec^{2} (x) - csc^{2} (x)

- tan (\frac{a}{2}) = - \frac{1}{csc ( a ) + cot ( a )}

6 sin^{2} (x) - 3 cos^{2} (x) = 9 sin^{2} (x) - 3

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

tan^{2} (θ) + cot^{2} (θ) = sec^{2} (θ) csc^{2} (θ)

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