Is (csc(θ)sin(θ))/(cos(θ))=sec(θ) ?

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

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prove (csc(θ)sin(θ))/(cos(θ))=sec(θ)

prove

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

True

Solution steps

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

Manipulating left side

\frac{csc ( θ ) sin ( θ )}{cos ( θ )}

Express with sin, cos

\frac{csc ( θ ) sin ( θ )}{cos ( θ )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply

\frac{1}{sin ( θ )} sin (θ) : 1

\frac{1}{sin ( θ )} sin (θ)

Multiply fractions:

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

= \frac{1 \cdot sin ( θ )}{sin ( θ )}

Cancel the common factor:

sin (θ)

= 1

= \frac{1}{cos ( θ )}

= \frac{1}{cos ( θ )}

= \frac{1}{cos ( θ )}

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{\frac{1}{s e c ( θ )}}

Simplify

\frac{1}{\frac{1}{s e c ( θ )}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{sec ( θ )}{1}

Apply rule

\frac{a}{1} = a

= sec (θ)

sec (θ)

sec (θ)

We showed that the two sides could take the same form

\Rightarrow True

cos (- \frac{π}{4}) = \frac{1}{2}

\frac{tan ( x ) + cot ( x )}{sec ( x ) csc ( x )} = 1

tan^{2} (a) + sin^{2} (a) = tan^{2} (a) sin^{2} (a)

tan (θ) sin (2 0^{\circ}) = 2 sin^{2} (θ)

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

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