Is (csc(θ)sin(θ))/(sec(θ))=cos(θ) ?

The answer to whether (csc(θ)sin(θ))/(sec(θ))=cos(θ) is True

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prove (csc(θ)sin(θ))/(sec(θ))=cos(θ)

prove

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

True

Solution steps

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

Manipulating left side

\frac{csc ( θ ) sin ( θ )}{sec ( θ )}

Express with sin, cos

\frac{csc ( θ ) sin ( θ )}{sec ( θ )}

Use the basic trigonometric identity:

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

= \frac{\frac{1}{s i n ( θ )} sin ( θ )}{sec ( θ )}

Use the basic trigonometric identity:

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

= \frac{\frac{1}{s i n ( θ )} sin ( θ )}{\frac{1}{c o s ( θ )}}

Simplify

\frac{\frac{1}{s i n ( θ )} sin ( θ )}{\frac{1}{c o s ( θ )}} : cos (θ)

\frac{\frac{1}{s i n ( θ )} sin ( θ )}{\frac{1}{c o s ( θ )}}

Apply the fraction rule:

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

= \frac{\frac{1}{s i n ( θ )} sin ( θ ) cos ( θ )}{1}

Apply the fraction rule:

\frac{a}{1} = a

= \frac{1}{sin ( θ )} sin (θ) cos (θ)

Multiply fractions:

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

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

Cancel the common factor:

sin (θ)

= 1 \cdot cos (θ)

Multiply:

1 \cdot cos (θ) = cos (θ)

= cos (θ)

= cos (θ)

= cos (θ)

We showed that the two sides could take the same form

\Rightarrow True

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

sin (x) - sin (x) \cdot cos^{2} (x) = sin^{3} (x)

(tan (x) + sec (x)) (1 - sin (x)) = cos (x)

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

cot (x) cos (2 x) = - sin (2 x) + cot (x)

Is (csc(θ)sin(θ))/(sec(θ))=cos(θ) ?
The answer to whether (csc(θ)sin(θ))/(sec(θ))=cos(θ) is True