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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (x)

= 1

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

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

= 1 - cos^{2} (x)

Rewrite using trig identities

1 - cos^{2} (x)

Use the Pythagorean identity:

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

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

= sin^{2} (x)

= sin^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

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

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

sec (x) + csc (x) = csc (x) (tan (x) + 1)

cos (θ + \frac{3 π}{2}) = sin (θ)

cot (x) \cdot sec (x) \cdot sin (x) = 1

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