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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove tan^2(x)+1= 1/(cos^2(x))

prove

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

True

Solution steps

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

Manipulating left side

tan^{2} (x) + 1

Express with sin, cos

1 + tan^{2} (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

Convert element to fraction:

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

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

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

Multiply:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

\frac{1 + tan ^{2} ( θ )}{sec ( θ )} = sec (θ)

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

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

cos (x) \cdot csc (x) = cot (x)

sin (4 x) = 4 sin (x) cos (x) cos (2 x)

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