Is ((tan^2(θ)))/((sec(θ)-1))-1=sec(θ) ?

The answer to whether ((tan^2(θ)))/((sec(θ)-1))-1=sec(θ) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

prove ((tan^2(θ)))/((sec(θ)-1))-1=sec(θ)

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Cancel

\frac{sec ^{2} ( θ ) - 1}{sec ( θ ) - 1} : sec (θ) + 1

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

Factor

sec^{2} (θ) - 1 : (sec (θ) + 1) (sec (θ) - 1)

sec^{2} (θ) - 1

Rewrite

1

1^{2}

= sec^{2} (θ) - 1^{2}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

sec^{2} (θ) - 1^{2} = (sec (θ) + 1) (sec (θ) - 1)

= (sec (θ) + 1) (sec (θ) - 1)

= \frac{( sec ( θ ) + 1 ) ( sec ( θ ) - 1 )}{sec ( θ ) - 1}

Cancel the common factor:

sec (θ) - 1

= sec (θ) + 1

= sec (θ) + 1 - 1

1 - 1 = 0

= sec (θ)

= sec (θ)

= sec (θ)

We showed that the two sides could take the same form

\Rightarrow True

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