Is cos(-θ)csc(-θ)tan(-θ)=1 ?

The answer to whether cos(-θ)csc(-θ)tan(-θ)=1 is True

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prove cos(-θ)csc(-θ)tan(-θ)=1

prove

cos (- θ) csc (- θ) tan (- θ) = 1

True

Solution steps

cos (- θ) csc (- θ) tan (- θ) = 1

Manipulating left side

cos (- θ) csc (- θ) tan (- θ)

Use the negative angle identity:

cos (- x) = cos (x)

= cos (θ) csc (- θ) tan (- θ)

Use the negative angle identity:

tan (- x) = - tan (x)

= cos (θ) csc (- θ) (- tan (θ))

Use the negative angle identity:

csc (- x) = - csc (x)

= cos (θ) (- csc (θ)) (- tan (θ))

Express with sin, cos

(- csc (θ)) (- tan (θ)) cos (θ)

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

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

Use the basic trigonometric identity:

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

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

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

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

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{1}{sin ( θ )} \cdot \frac{sin ( θ )}{cos ( θ )} cos (θ)

Multiply fractions:

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

= \frac{1 \cdot sin ( θ ) cos ( θ )}{sin ( θ ) cos ( θ )}

Cancel the common factor:

sin (θ)

= \frac{1 \cdot cos ( θ )}{cos ( θ )}

Cancel the common factor:

cos (θ)

= 1

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

p ro v e sec (x) - 2 tan (x) = 0

p ro v e cos (3 θ) = cos (θ) (cos^{2} (θ) - 3 sin^{2} (θ))

p ro v e cos (θ) = sec (θ) - (sin (θ)) (tan (θ))

p ro v e cos (2 x - 90) = 2 sin (x) cos (x)

p ro v e sin^{4} (x) - 2 sin^{2} (x) = cos^{4} (x) - 1

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