Is (tan(θ))/(cot(θ))=tan^2(θ) ?

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

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prove (tan(θ))/(cot(θ))=tan^2(θ)

prove

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

True

Solution steps

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

Manipulating left side

\frac{tan ( θ )}{cot ( θ )}

Express with sin, cos

\frac{tan ( θ )}{cot ( θ )}

Use the basic trigonometric identity:

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

= \frac{\frac{s i n ( θ )}{c o s ( θ )}}{cot ( θ )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

Divide fractions:

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

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

sin (θ) sin (θ) = sin^{2} (θ)

sin (θ) sin (θ)

Apply exponent rule:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= sin^{2} (θ)

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

cos (θ) cos (θ) = cos^{2} (θ)

cos (θ) cos (θ)

Apply exponent rule:

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

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= cos^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= cos^{2} (θ)

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

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

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

\frac{sin ( θ ) tan ( θ )}{cos ( θ )}

= tan (θ) \frac{sin ( θ )}{cos ( θ )}

Use the basic trigonometric identity:

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

tan (θ) tan (θ)

Simplify

tan (θ) tan (θ) : tan^{2} (θ)

tan (θ) tan (θ)

Apply exponent rule:

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

tan (θ) tan (θ) = tan^{1 + 1} (θ)

= tan^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= tan^{2} (θ)

tan^{2} (θ)

tan^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

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

\frac{sin ( 6 x )}{cos ( x )} = 6 sin (x)

csc (x) sec (x) - tan (x) = cot (x)

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

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

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