Is (tan(x))/(csc(x))=sec(x)-cos(x) ?

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

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prove (tan(x))/(csc(x))=sec(x)-cos(x)

prove

\frac{tan ( x )}{csc ( x )} = sec (x) - cos (x)

True

Solution steps

\frac{tan ( x )}{csc ( x )} = sec (x) - cos (x)

Manipulating left side

\frac{tan ( x )}{csc ( x )}

Express with sin, cos

\frac{tan ( x )}{csc ( x )}

Use the basic trigonometric identity:

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

= \frac{\frac{s i n ( x )}{c o s ( x )}}{csc ( x )}

Use the basic trigonometric identity:

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

= \frac{\frac{s i n ( x )}{c o s ( x )}}{\frac{1}{s i n ( x )}}

Simplify

\frac{\frac{s i n ( x )}{c o s ( x )}}{\frac{1}{s i n ( x )}} : \frac{sin ^{2} ( x )}{cos ( x )}

\frac{\frac{s i n ( x )}{c o s ( x )}}{\frac{1}{s i n ( x )}}

Divide fractions:

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

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

Refine

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

sin (x) sin (x) = sin^{2} (x)

sin (x) sin (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

sin (x) sin (x) = sin^{1 + 1} (x)

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

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

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

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

Rewrite using trig identities

\frac{sin ^{2} ( x )}{cos ( x )}

Use the Pythagorean identity:

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

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

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

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

Manipulating right side

sec (x) - cos (x)

Express with sin, cos

- cos (x) + sec (x)

Use the basic trigonometric identity:

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

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

Simplify

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

- cos (x) + \frac{1}{cos ( x )}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

- cos (x) cos (x) + 1

cos (x) cos (x) = cos^{2} (x)

cos (x) cos (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cos (x) cos (x) = cos^{1 + 1} (x)

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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