Is 1+tan^2(a)= 1/(cos^2(a)) ?

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

prove

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

True

Solution steps

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

Manipulating left side

1 + tan^{2} (a)

Express with sin, cos

1 + tan^{2} (a)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

= 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 )}

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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

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

cos (x + 1) = cos (- x - 1)

cos (x) (tan (x) sec (x)) = sin (x) - 1

\frac{tan ( θ )}{sin ( θ )} = sec (θ)

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