Is 1/(tan(b))+tan(b)=(sec^2(b))/(tan(b)) ?

The answer to whether 1/(tan(b))+tan(b)=(sec^2(b))/(tan(b)) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

prove 1/(tan(b))+tan(b)=(sec^2(b))/(tan(b))

prove

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

True

Solution steps

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

Manipulating left side

\frac{1}{tan ( b )} + tan (b)

Simplify

\frac{1}{tan ( b )} + tan (b) : \frac{1 + tan ^{2} ( b )}{tan ( b )}

\frac{1}{tan ( b )} + tan (b)

Convert element to fraction:

tan (b) = \frac{tan ( b ) tan ( b )}{tan ( b )}

= \frac{1}{tan ( b )} + \frac{tan ( b ) tan ( b )}{tan ( b )}

Since the denominators are equal, combine the fractions:

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

= \frac{1 + tan ( b ) tan ( b )}{tan ( b )}

1 + tan (b) tan (b) = 1 + tan^{2} (b)

1 + tan (b) tan (b)

tan (b) tan (b) = tan^{2} (b)

tan (b) tan (b)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

tan (b) tan (b) = tan^{1 + 1} (b)

= tan^{1 + 1} (b)

Add the numbers:

1 + 1 = 2

= tan^{2} (b)

= 1 + tan^{2} (b)

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

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

Rewrite using trig identities

\frac{1 + tan ^{2} ( b )}{tan ( b )}

Use the Pythagorean identity:

tan^{2} (x) + 1 = sec^{2} (x)

= \frac{sec ^{2} ( b )}{tan ( b )}

= \frac{sec ^{2} ( b )}{tan ( b )}

We showed that the two sides could take the same form

\Rightarrow True

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