Is cos(θ)tan(-θ)=-sin(θ) ?

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

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prove cos(θ)tan(-θ)=-sin(θ)

prove

cos (θ) tan (- θ) = - sin (θ)

True

Solution steps

cos (θ) tan (- θ) = - sin (θ)

Manipulating left side

cos (θ) tan (- θ)

Use the negative angle identity:

tan (- x) = - tan (x)

= cos (θ) (- tan (θ))

Express with sin, cos

(- tan (θ)) cos (θ)

Use the basic trigonometric identity:

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

= (- \frac{sin ( θ )}{cos ( θ )}) cos (θ)

Simplify

(- \frac{sin ( θ )}{cos ( θ )}) cos (θ) : - sin (θ)

(- \frac{sin ( θ )}{cos ( θ )}) cos (θ)

Remove parentheses:

(- a) = - a

= - \frac{sin ( θ )}{cos ( θ )} cos (θ)

Multiply fractions:

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

= - \frac{sin ( θ ) cos ( θ )}{cos ( θ )}

Cancel the common factor:

cos (θ)

= - sin (θ)

= - sin (θ)

= - sin (θ)

We showed that the two sides could take the same form

\Rightarrow True

tan^{3} (x) cot (x) = sec^{2} (x) - 1

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

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

sec (θ) = \frac{tan ( θ )}{csc ( θ )} + cos (θ)

sin (\frac{7 π}{6}) = - \frac{1}{2}