Is 1/(csc(x)-sin(x))=tan(x)sec(x) ?

The answer to whether 1/(csc(x)-sin(x))=tan(x)sec(x) is True

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prove 1/(csc(x)-sin(x))=tan(x)sec(x)

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Manipulating right side

tan (x) sec (x)

Express with sin, cos

sec (x) tan (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply:

1 \cdot sin (x) = sin (x)

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

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)

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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Is 1/(csc(x)-sin(x))=tan(x)sec(x) ?
The answer to whether 1/(csc(x)-sin(x))=tan(x)sec(x) is True