Is sin(θ)csc(θ)-cos^2(θ)=sin^2(θ) ?

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

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prove sin(θ)csc(θ)-cos^2(θ)=sin^2(θ)

prove

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

True

Solution steps

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

Manipulating left side

sin (θ) csc (θ) - cos^{2} (θ)

Express with sin, cos

- cos^{2} (θ) + csc (θ) sin (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

\frac{1}{sin ( θ )} sin (θ) = 1

\frac{1}{sin ( θ )} sin (θ)

Multiply fractions:

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

= \frac{1 \cdot sin ( θ )}{sin ( θ )}

Cancel the common factor:

sin (θ)

= 1

= - cos^{2} (θ) + 1

= - cos^{2} (θ) + 1

= 1 - cos^{2} (θ)

Rewrite using trig identities

1 - cos^{2} (θ)

Use the Pythagorean identity:

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

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

= sin^{2} (θ)

= sin^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

tan^{2} (θ) + 1 = \frac{1}{cos ^{2} ( θ )}

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

\frac{2 tan ( x )}{sec ^{2} ( x )} = sin (2 x)

csc^{2} (x) - 1 = cot^{2} (x)

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

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