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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

prove

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

True

Solution steps

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

Manipulating left side

(1 + tan^{2} (u)) (1 - sin^{2} (u))

Rewrite using trig identities

(1 + tan^{2} (u)) (1 - sin^{2} (u))

Use the Pythagorean identity:

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

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

= (1 + tan^{2} (u)) cos^{2} (u)

= (1 + tan^{2} (u)) cos^{2} (u)

Express with sin, cos

(1 + tan^{2} (u)) cos^{2} (u)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (u)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

sec^{3} (x) = \frac{1}{cos ^{3} ( x )}

sin (x) = \frac{5}{13}

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

sin^{4} (x) + cos^{4} (x) = 1 - \frac{1}{2} sin^{2} (2 x)

\frac{( 1 + cos ( 2 y ) )}{sin ^{2} ( y )} = 2 cot^{2} (y)

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