Is (1+tan^2(θ))/(1+cot^2(θ))=tan^2(θ) ?

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

Apply exponent rule:

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

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot sin^{2} (θ) = sin^{2} (θ)

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

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot cos^{2} (θ) = cos^{2} (θ)

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

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

Cancel the common factor:

cos^{2} (θ) + sin^{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

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Frequently Asked Questions (FAQ)

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