What is the general solution for 1+(tan^2(θ))/(sec(θ))=sec(θ) ?

The general solution for 1+(tan^2(θ))/(sec(θ))=sec(θ) is θ=2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

1+(tan^2(θ))/(sec(θ))=sec(θ)

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

θ = 2 πn

Solution steps

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

Subtract

sec (θ)

from both sides

1 + \frac{tan ^{2} ( θ )}{sec ( θ )} - sec (θ) = 0

Simplify

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

\frac{sec ( θ ) + tan ^{2} ( θ ) - sec ^{2} ( θ )}{sec ( θ )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

sec (θ) + tan^{2} (θ) - sec^{2} (θ) = 0

Rewrite using trig identities

sec (θ) - 1 = 0

Move

1

to the right side

sec (θ) = 1

General solutions for

sec (θ) = 1

θ = 0 + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 2 πn

170 = \frac{16 0 ^{2}}{32} sin (2 θ)

sin (3 x) = 3 sin (x) cos (x)

sin (x) = \frac{2 π}{5}

sin (x) = \frac{2 π}{7}

cos (θ) sin (θ) = 0

What is the general solution for 1+(tan^2(θ))/(sec(θ))=sec(θ) ?
The general solution for 1+(tan^2(θ))/(sec(θ))=sec(θ) is θ=2pin