Is tan(θ)=(sec(θ))/(csc(θ)) ?

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

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prove tan(θ)=(sec(θ))/(csc(θ))

prove

tan (θ) = \frac{sec ( θ )}{csc ( θ )}

True

Solution steps

tan (θ) = \frac{sec ( θ )}{csc ( θ )}

Manipulating left side

tan (θ)

Express with sin, cos

tan (θ)

Use the basic trigonometric identity:

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

= \frac{sin ( θ )}{cos ( θ )}

= \frac{sin ( θ )}{cos ( θ )}

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{\frac{1}{c s c ( θ )}}{cos ( θ )}

Use the basic trigonometric identity:

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

\frac{\frac{1}{c s c ( θ )}}{\frac{1}{s e c ( θ )}}

Simplify

\frac{\frac{1}{c s c ( θ )}}{\frac{1}{s e c ( θ )}}

Divide fractions:

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

= \frac{1 \cdot sec ( θ )}{csc ( θ ) \cdot 1}

Refine

= \frac{sec ( θ )}{csc ( θ )}

\frac{sec ( θ )}{csc ( θ )}

\frac{sec ( θ )}{csc ( θ )}

We showed that the two sides could take the same form

\Rightarrow True

\frac{sec ( θ )}{csc ^{2} ( θ )} = sec (θ) - cos (θ)

cos (3 β) = cos (β) (cos^{2} (β) - 3 sin^{2} (β))

sec^{2} (θ) cot^{2} (θ) - 1 = cot^{2} (θ)

1 - \frac{( sin ^{2} ( x ) )}{1 + cos ( x )} = cos (x)

cot (a) = \frac{1}{tan ( a )}

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