Is sin(θ)cos(θ)tan(θ)=1-cos^2(θ) ?

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

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prove sin(θ)cos(θ)tan(θ)=1-cos^2(θ)

prove

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

True

Solution steps

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

Manipulating left side

sin (θ) cos (θ) tan (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)

Rewrite using trig identities

sin^{2} (θ)

Use the Pythagorean identity:

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

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

= 1 - cos^{2} (θ)

= 1 - cos^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

cot^{2} (- θ) + 1 = csc^{2} (θ)

\frac{1 - csc ^{2} ( t )}{csc ^{2} ( t )} = \frac{- 1}{sec ^{2} ( t )}

cot (θ) = cos (θ) csc (θ)

sech^{2} (x) = 1 - tanh^{2} (x)

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

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