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

The answer to whether (1-sin^2(a))(1+tan^2(a))=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-sin^2(a))(1+tan^2(a))=1

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

Join

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

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

Convert element to fraction:

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

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

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

Multiply:

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (a)

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

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

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

Rewrite using trig identities

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

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

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

tan (x) = \frac{1}{cot ( x )}

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

cot (t) csc (t) = \frac{csc ^{2} ( t ) - 1}{cos ( t )}

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

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