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

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

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

prove

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

True

Solution steps

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

Manipulating left side

sin (x) tan (x) cos (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{sin ( θ ) + sin ( 3 θ )}{2 sin ( 2 θ )} = cos (θ)

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

cos (b) csc (b) tan (b) = 1

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

tan^{6} (x) = tan^{4} (x) sec^{2} (x) - tan^{4} (x)

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