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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

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

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

Multiply

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

\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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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

(sec (a) - tan (a)) (sec (a) + tan (a)) = 1

sec (x - \frac{π}{2}) = csc (x)

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

tan (x) (1 - csc^{2} (x)) = - cot (x)

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