Is 1-(cot^2(θ))/(csc^2(θ))=sin^2(θ) ?

The answer to whether 1-(cot^2(θ))/(csc^2(θ))=sin^2(θ) is True

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prove 1-(cot^2(θ))/(csc^2(θ))=sin^2(θ)

prove

1 - \frac{cot ^{2} ( θ )}{csc ^{2} ( θ )} = sin^{2} (θ)

True

Solution steps

1 - \frac{cot ^{2} ( θ )}{csc ^{2} ( θ )} = sin^{2} (θ)

Manipulating left side

1 - \frac{cot ^{2} ( θ )}{csc ^{2} ( θ )}

Express with sin, cos

1 - \frac{cot ^{2} ( θ )}{csc ^{2} ( θ )}

Use the basic trigonometric identity:

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

= 1 - \frac{( \frac{c o s ( θ )}{s i n ( θ )} ) ^{2}}{csc ^{2} ( θ )}

Use the basic trigonometric identity:

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

= 1 - \frac{( \frac{c o s ( θ )}{s i n ( θ )} ) ^{2}}{( \frac{1}{s i n ( θ )} ) ^{2}}

Simplify

1 - \frac{( \frac{c o s ( θ )}{s i n ( θ )} ) ^{2}}{( \frac{1}{s i n ( θ )} ) ^{2}} : 1 - cos^{2} (θ)

1 - \frac{( \frac{c o s ( θ )}{s i n ( θ )} ) ^{2}}{( \frac{1}{s i n ( θ )} ) ^{2}}

\frac{( \frac{c o s ( θ )}{s i n ( θ )} ) ^{2}}{( \frac{1}{s i n ( θ )} ) ^{2}} = cos^{2} (θ)

\frac{( \frac{c o s ( θ )}{s i n ( θ )} ) ^{2}}{( \frac{1}{s i n ( θ )} ) ^{2}}

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

(\frac{1}{sin ( θ )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{sin ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{sin ^{2} ( θ )}

= \frac{( \frac{c o s ( θ )}{s i n ( θ )} ) ^{2}}{\frac{1}{s i n ^{2} ( θ )}}

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

(\frac{cos ( θ )}{sin ( θ )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

= \frac{\frac{c o s ^{2} ( θ )}{s i n ^{2} ( θ )}}{\frac{1}{s i n ^{2} ( θ )}}

Divide fractions:

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

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

Refine

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

Cancel the common factor:

sin^{2} (θ)

= cos^{2} (θ)

= 1 - cos^{2} (θ)

= 1 - cos^{2} (θ)

= 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

(40 \cdot cos (θ))^{2} + (40 \cdot sin (θ))^{2} = 4 0^{2}

1 = sec^{2} (x)

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

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

csc^{2} (t) tan^{2} (t) - 1 = tan^{2} (t)

Is 1-(cot^2(θ))/(csc^2(θ))=sin^2(θ) ?
The answer to whether 1-(cot^2(θ))/(csc^2(θ))=sin^2(θ) is True