Is (csc^2(θ))/(1+tan^2(θ))=cot^2(θ) ?

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

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prove (csc^2(θ))/(1+tan^2(θ))=cot^2(θ)

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

tan^{2} (x) + 1 = sec^{2} (x)

= \frac{csc ^{2} ( θ )}{sec ^{2} ( θ )}

= \frac{csc ^{2} ( θ )}{sec ^{2} ( θ )}

Express with sin, cos

\frac{csc ^{2} ( θ )}{sec ^{2} ( θ )}

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

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

Simplify

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

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

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

(\frac{1}{cos ( θ )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{cos ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( θ )}

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

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

Divide fractions:

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

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

Refine

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

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

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

Rewrite using trig identities

= \frac{cos ( θ )}{sin ( θ )} \cdot \frac{cos ( θ )}{sin ( θ )}

Use the basic trigonometric identity:

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

\frac{cos ( θ ) cot ( θ )}{sin ( θ )}

= cot (θ) \frac{cos ( θ )}{sin ( θ )}

Use the basic trigonometric identity:

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

cot (θ) cot (θ)

Simplify

cot (θ) cot (θ) : cot^{2} (θ)

cot (θ) cot (θ)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cot (θ) cot (θ) = cot^{1 + 1} (θ)

= cot^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= cot^{2} (θ)

cot^{2} (θ)

cot^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

\frac{tan ( y )}{csc ( y )} = sec (y) - cos (y)

cos (3 x) = cos (2 x + x)

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

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

1 - sin^{2} (θ) = cos^{2} (θ)

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