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

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

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

prove

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

True

Solution steps

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

Manipulating left side

csc (x) sin (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

\frac{tan ^{2} ( A )}{tan ^{2} ( A ) + 1} = sin^{2} (A)

\frac{sec ^{4} ( α ) - 1}{tan ^{2} ( α )} = sec^{2} (α) + 1

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

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

cos (8 x) = cos^{2} (4 x) - sin^{2} (4 x)

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