Is tan(x)csc(x)=sec(x) ?

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

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prove tan(x)csc(x)=sec(x)

prove

tan (x) csc (x) = sec (x)

True

Solution steps

tan (x) csc (x) = sec (x)

Manipulating left side

tan (x) csc (x)

Express with sin, cos

csc (x) tan (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (x)

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

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

Simplify

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

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= sec (x)

sec (x)

sec (x)

We showed that the two sides could take the same form

\Rightarrow True

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

tan (u) - sec (u) = - \frac{cos ( u )}{1 + sin ( u )}

\frac{csc ( - x )}{sec ( - x )} = - cot (x)

(1 + tan^{2} (θ)) cos^{2} (θ) = 1

cos (x - y) + cos (x + y) = 2 cos (x) cos (y)