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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

- sin^{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

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

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

= 1 - sin^{2} (x)

Rewrite using trig identities

1 - sin^{2} (x)

Use the Pythagorean identity:

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

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

= cos^{2} (x)

= cos^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

tan (2 π + θ) = tan (θ)

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

cos (a) + sin (a) tan (a) = sec (a)

(csc (θ) + cot (θ)) (1 - cos (θ)) = sin (θ)

\frac{1 - cos ( x )}{sin ( x )} = csc (x) - cot (x)

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