Is sin(-θ)sec(-θ)cot(-θ)=1 ?

The answer to whether sin(-θ)sec(-θ)cot(-θ)=1 is True

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prove sin(-θ)sec(-θ)cot(-θ)=1

prove

sin (- θ) sec (- θ) cot (- θ) = 1

True

Solution steps

sin (- θ) sec (- θ) cot (- θ) = 1

Manipulating left side

sin (- θ) sec (- θ) cot (- θ)

Use the negative angle identity:

sin (- x) = - sin (x)

= cot (- θ) sec (- θ) (- sin (θ))

Use the negative angle identity:

sec (- x) = sec (x)

= cot (- θ) sec (θ) (- sin (θ))

Use the negative angle identity:

cot (- x) = - cot (x)

= (- cot (θ)) sec (θ) (- sin (θ))

Express with sin, cos

(- cot (θ)) (- sin (θ)) sec (θ)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

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

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

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

Remove parentheses:

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (θ)

= \frac{1 \cdot sin ( θ )}{sin ( θ )}

Cancel the common factor:

sin (θ)

= 1

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

cos^{4} (θ) - sin^{4} (θ) = 1 - 2 sin^{2} (θ)

cot (x) sin (x) sec^{2} (x) = sec (x)

tan (2 x) = \frac{sin ( 2 x )}{cos ( 2 x )}

sin^{2} (x) + 4 cos^{2} (x) = 4 - 3 sin^{2} (x)

sin (a + b) = sin (a) cos (b) + cos (a) sin (b)

Is sin(-θ)sec(-θ)cot(-θ)=1 ?
The answer to whether sin(-θ)sec(-θ)cot(-θ)=1 is True