Is csc(θ)tan(θ)=sec(θ) ?

The answer to whether csc(θ)tan(θ)=sec(θ) is True

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prove csc(θ)tan(θ)=sec(θ)

prove

csc (θ) tan (θ) = sec (θ)

True

Solution steps

csc (θ) tan (θ) = sec (θ)

Manipulating left side

csc (θ) tan (θ)

Express with sin, cos

csc (θ) tan (θ)

Use the basic trigonometric identity:

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

= \frac{1}{sin ( θ )} tan (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (θ)

= \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

p ro v e \frac{1 + sin ( - x )}{1 + csc ( - x )} = - sin (x)

p ro v e tan^{2} (x) + 1 = sec^{2} (x)

p ro v e cos (- x) = cos (x)

p ro v e cosh^{2} (x) - sinh^{2} (x) = 1

p ro v e \frac{1 - tan ^{2} ( x )}{1 + tan ^{2} ( x )} = cos (2 x)