Is (1+tan^2(u))/(1-tan^2(u))= 1/(cos^2(u)-sin^2(u)) ?

The answer to whether (1+tan^2(u))/(1-tan^2(u))= 1/(cos^2(u)-sin^2(u)) is True

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prove (1+tan^2(u))/(1-tan^2(u))= 1/(cos^2(u)-sin^2(u))

prove

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

True

Solution steps

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

Manipulating left side

\frac{1 + tan ^{2} ( u )}{1 - tan ^{2} ( u )}

Express with sin, cos

\frac{1 + tan ^{2} ( u )}{1 - tan ^{2} ( u )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

Apply exponent rule:

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

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

Join

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

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

Convert element to fraction:

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

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

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} ( u ) - sin ^{2} ( u )}{cos ^{2} ( u )}

Multiply:

1 \cdot cos^{2} (u) = cos^{2} (u)

= \frac{cos ^{2} ( u ) - sin ^{2} ( u )}{cos ^{2} ( u )}

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

Join

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

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

Convert element to fraction:

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

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

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} ( u ) + sin ^{2} ( u )}{cos ^{2} ( u )}

Multiply:

1 \cdot cos^{2} (u) = cos^{2} (u)

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

= \frac{\frac{c o s ^{2} ( u ) + s i n ^{2} ( u )}{c o s ^{2} ( u )}}{\frac{c o s ^{2} ( u ) - s i n ^{2} ( u )}{c o s ^{2} ( u )}}

Divide fractions:

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

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

Cancel the common factor:

cos^{2} (u)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

csc (θ) + cot (θ) = \frac{sin ( θ )}{1 - cos ( θ )}

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

(csc (x) + cot (x)) (1 - cos (x)) = sin (x)

tan (\frac{π}{2} - x) = cot (x)

sec (θ) = \frac{tan ( θ )}{sin ( θ )}

Is (1+tan^2(u))/(1-tan^2(u))= 1/(cos^2(u)-sin^2(u)) ?
The answer to whether (1+tan^2(u))/(1-tan^2(u))= 1/(cos^2(u)-sin^2(u)) is True