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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply the fraction rule:

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

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

Apply exponent rule:

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

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

Join

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

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

Convert element to fraction:

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

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

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} ( a ) + cos ^{2} ( a )}{sin ^{2} ( a )}

Multiply:

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

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (a)

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

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

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

Apply rule

\frac{a}{1} = a

= sin (a)

= sin (a)

We showed that the two sides could take the same form

\Rightarrow True

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

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

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

csc (x) = 2

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

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