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

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

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

prove

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

True

Solution steps

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

Manipulating left side

csc (x) tan (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

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

(sin (x)) (csc (x)) = 1

1 + cos (2 t) = cot (t) sin (2 t)

sin (α) = \frac{2 tan ( \frac{α}{2} )}{1 + tan ^{2} ( \frac{α}{2} )}

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

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