Is 1+tan^2(α)= 1/(cos^2(α)) ?

The answer to whether 1+tan^2(α)= 1/(cos^2(α)) 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(α)= 1/(cos^2(α))

prove

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

True

Solution steps

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

Manipulating left side

1 + tan^{2} (α)

Express with sin, cos

1 + tan^{2} (α)

Use the basic trigonometric identity:

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

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

Simplify

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

1 + (\frac{sin ( α )}{cos ( α )})^{2}

Apply exponent rule:

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

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

Convert element to fraction:

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

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

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

Multiply:

1 \cdot cos^{2} (α) = cos^{2} (α)

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

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

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

Rewrite using trig identities

\frac{cos ^{2} ( α ) + sin ^{2} ( α )}{cos ^{2} ( α )}

Use the Pythagorean identity:

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

= \frac{1}{cos ^{2} ( α )}

= \frac{1}{cos ^{2} ( α )}

We showed that the two sides could take the same form

\Rightarrow True

sin (x) (cot (x) + tan (x)) = sec (x)

\frac{tan ( x ) - cot ( x )}{tan ( x ) + cot ( x )} = 1 - 2 cos^{2} (x)

tan (θ) cos (θ) = sin (θ)

\frac{sin ( x + \frac{π}{2} )}{sin ( π - x )} = cot (x)

tan (θ) + \frac{cos ( θ )}{1 + sin ( θ )} = sec (θ)

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