Is tan(x)(csc(x)-sin(x))=cos(x) ?

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

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prove tan(x)(csc(x)-sin(x))=cos(x)

prove

tan (x) (csc (x) - sin (x)) = cos (x)

True

Solution steps

tan (x) (csc (x) - sin (x)) = cos (x)

Manipulating left side

tan (x) (csc (x) - sin (x))

Express with sin, cos

(csc (x) - sin (x)) tan (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

= \frac{sin ( x ) ( \frac{1}{s i n ( x )} - sin ( x ) )}{cos ( x )}

Join

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

\frac{1}{sin ( x )} - sin (x)

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

1 - sin (x) sin (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)

= 1 - sin^{2} (x)

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (x)

= 1 - sin^{2} (x)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

cos (x)

= cos (x)

= cos (x)

We showed that the two sides could take the same form

\Rightarrow True

tan (x) (sin (x) + cot (x) cos (x)) = sec (x)

sin (x) cos^{2} (x) + sin^{3} (x) = sin (x)

\frac{csc ( x ) + sec ( x )}{tan ( x ) + 1} = csc (x)

cos (\frac{π}{3}) = \frac{1}{2}

sec (- x) = sec (x)

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