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

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

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prove 1+1/(tan^2(x))=csc^2(x)

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

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

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

Apply exponent rule:

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

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

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

Apply the fraction rule:

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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