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

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

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

prove

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

True

Solution steps

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

Manipulating left side

sin^{2} (x) + sin^{2} (x) cot^{2} (x)

Express with sin, cos

sin^{2} (x) + cot^{2} (x) sin^{2} (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin^{2} (x)

= cos^{2} (x)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

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

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

cot (t) csc (t) = \frac{csc ^{2} ( t ) - 1}{cos ( t )}

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

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

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