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

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

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prove (csc(x))/(1+cot^2(x))=sin(x)

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply the fraction rule:

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

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

Apply exponent rule:

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

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

Join

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

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

Multiply:

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

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (x)

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

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

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

Apply rule

\frac{a}{1} = a

= sin (x)

= sin (x)

We showed that the two sides could take the same form

\Rightarrow True

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

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

sec (\frac{3 π}{2} - x) = - csc (x)

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

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

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