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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

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

sin^{2} (2 x) = \frac{1}{2} (1 - cos (4 x))

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

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

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