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

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1+(tan^2(θ))/(sec(θ))=sec(θ)

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

θ = 2 πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n

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 )}

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

Convert element to fraction:

1 = \frac{1 sec ( θ )}{sec ( θ )}, sec (θ) = \frac{sec ( θ ) sec ( θ )}{sec ( θ )}

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

Since the denominators are equal, combine the fractions:

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

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

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

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

1 \cdot sec (θ) = sec (θ)

1 \cdot sec (θ)

Multiply:

1 \cdot sec (θ) = sec (θ)

= sec (θ)

sec (θ) sec (θ) = sec^{2} (θ)

sec (θ) sec (θ)

Apply exponent rule:

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

sec (θ) sec (θ) = sec^{1 + 1} (θ)

= sec^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= sec^{2} (θ)

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

Use the Pythagorean identity:

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

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

= sec (θ) - 1

sec (θ) - 1 = 0

Move

1

to the right side

sec (θ) - 1 = 0

Add

1

to both sides

sec (θ) - 1 + 1 = 0 + 1

Simplify

sec (θ) = 1

sec (θ) = 1

General solutions for

sec (θ) = 1

sec (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

θ = 0 + 2 πn

θ = 0 + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 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