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

Popular >

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

= sec (θ)

We showed that the two sides could take the same form

\Rightarrow True

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

tan (x) = cot (x) - 2 cot (2 x)

tan (3 x) (tan (x) - 1) = 0

cos (x) + sin (x) tan^{2} (x) = sec (x)

csc^{2} (x) - cot^{2} (x) + 1 = 0

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