Is sin(u)csc(u)-cos^2(u)=sin^2(u) ?

The answer to whether sin(u)csc(u)-cos^2(u)=sin^2(u) is True

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prove sin(u)csc(u)-cos^2(u)=sin^2(u)

prove

sin (u) csc (u) - cos^{2} (u) = sin^{2} (u)

True

Solution steps

sin (u) csc (u) - cos^{2} (u) = sin^{2} (u)

Manipulating left side

sin (u) csc (u) - cos^{2} (u)

Express with sin, cos

- cos^{2} (u) + csc (u) sin (u)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

\frac{1}{sin ( u )} sin (u)

Multiply fractions:

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

= \frac{1 \cdot sin ( u )}{sin ( u )}

Cancel the common factor:

sin (u)

= 1

= - cos^{2} (u) + 1

= - cos^{2} (u) + 1

= 1 - cos^{2} (u)

Rewrite using trig identities

1 - cos^{2} (u)

Use the Pythagorean identity:

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

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

= sin^{2} (u)

= sin^{2} (u)

We showed that the two sides could take the same form

\Rightarrow True

sin (x) + cot (x) cos (x) = csc (x)

3 sin^{2} (θ) + 4 cos^{2} (θ) = 3 + cos^{2} (θ)

(1 + cot^{2} (θ)) sin^{2} (θ) = 1

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

sin^{2} (- θ) + cos^{2} (- θ) = 1

Is sin(u)csc(u)-cos^2(u)=sin^2(u) ?
The answer to whether sin(u)csc(u)-cos^2(u)=sin^2(u) is True