Is 1/(sec(θ))=(sec(θ))^{-1} ?

The answer to whether 1/(sec(θ))=(sec(θ))^{-1} is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove 1/(sec(θ))=(sec(θ))^{-1}

prove

\frac{1}{sec ( θ )} = (sec (θ))^{- 1}

True

Solution steps

\frac{1}{sec ( θ )} = (sec (θ))^{- 1}

Manipulating right side

(sec (θ))^{- 1}

Simplify

= sec^{- 1} (θ)

Simplify

= \frac{1}{sec ( θ )}

We showed that the two sides could take the same form

\Rightarrow True

sin^{2} (2 x) = \frac{1}{2 ( 1 - cos ( 4 x ) )}

sin (2 θ) + cos (2 θ) = 1

25 (sec^{2} (5 x) - tan^{2} (5 x)) = 25

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

+ \frac{cos ( x )}{1 - sin ( x )} = sec (x) + tan (x)

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