Is (1-sin(x))/(1-1/(sin(x)))=-sin(x) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove (1-sin(x))/(1-1/(sin(x)))=-sin(x)

prove

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

True

Solution steps

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

Manipulating left side

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

Simplify

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

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

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

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

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

Apply the fraction rule:

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

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

sin (x) - 1 = - (- sin (x) + 1)

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

Refine

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

Cancel the common factor:

1 - sin (x)

= - sin (x)

= - sin (x)

We showed that the two sides could take the same form

\Rightarrow True

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

\frac{cos ( t )}{1 + sin ( t )} + \frac{1 - sin ( t )}{cos ( t )} = 2 (sec (t) - tan (t))

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

\frac{cot ( 2 θ )}{cos ( 2 θ )} = csc (2 θ)

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

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