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

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

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

prove

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

True

Solution steps

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

Manipulating left side

1 + cos^{2} (x) csc^{2} (x)

Express with sin, cos

1 + cos^{2} (x) csc^{2} (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Multiply fractions:

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

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

Multiply:

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

= \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 )}

We showed that the two sides could take the same form

\Rightarrow True

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

sin (x) cos (x) = sin (2 x)

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

tan^{2} (\frac{5 π}{4}) = tan (\frac{5 π}{4})

sin (\frac{π}{2} - θ) = sin (θ - \frac{π}{2})

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