Is (1+csc^2(x))/(csc^2(x))=1+sin^2(x) ?

The answer to whether (1+csc^2(x))/(csc^2(x))=1+sin^2(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+csc^2(x))/(csc^2(x))=1+sin^2(x)

prove

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

True

Solution steps

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

Manipulating left side

\frac{1 + csc ^{2} ( x )}{csc ^{2} ( x )}

Express with sin, cos

\frac{1 + csc ^{2} ( x )}{csc ^{2} ( x )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

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

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Apply the fraction rule:

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

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

Join

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

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

Convert element to fraction:

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

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

Multiply:

1 \cdot sin^{2} (x) = sin^{2} (x)

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

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

Apply the fraction rule:

\frac{a}{1} = a

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

Multiply fractions:

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

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

Cancel the common factor:

sin^{2} (x)

= sin^{2} (x) + 1

= sin^{2} (x) + 1

= 1 + sin^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

(sin (a) + cos (a))^{2} - 2 sin (a) cos (a) = 1

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

81 - 9 (3 sin (x))^{2} = 4 sin (x) + 8

cot (x) \cdot cos (x) + sin (x) = csc (x)

(1 - cos^{2} (a)) \cdot (1 + tan^{2} (a)) = tan^{2} (a)

Is (1+csc^2(x))/(csc^2(x))=1+sin^2(x) ?
The answer to whether (1+csc^2(x))/(csc^2(x))=1+sin^2(x) is True