Is (4sec(x)-4)/(1-cos(θ))=4sec(x) ?

The answer to whether (4sec(x)-4)/(1-cos(θ))=4sec(x) is False

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prove (4sec(x)-4)/(1-cos(θ))=4sec(x)

prove

\frac{4 sec ( x ) - 4}{1 - cos ( θ )} = 4 sec (x)

False

Solution steps

\frac{4 sec ( x ) - 4}{1 - cos ( θ )} = 4 sec (x)

Show that the two sides are not equal

For

\frac{4 sec ( x ) - 4}{1 - cos ( θ )} = 4 sec (x)

plug in

x = 0

For

\frac{4 sec ( x ) - 4}{1 - cos ( θ )} = 4 sec (x)

plug in

θ = 1

\frac{4 sec ( 0 ) - 4}{1 - cos ( 1 )} = 0

\frac{4 sec ( 0 ) - 4}{1 - cos ( 1 )}

Refine to a decimal form

= 0

4 sec (0) = 4

4 sec (0)

Use the following trivial identity:

sec (0) = 1

sec (0)

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}

= 1

= 4 \cdot 1

Simplify

= 4

The two sides are not equal

\Rightarrow False

p ro v e 1 + cot^{2} (x) = \frac{tan ^{2} ( x ) + 1}{tan ^{2} ( x )}

p ro v e sin (18 0^{\circ} + θ) = - sin (θ)

p ro v e \frac{1 - tan ^{2} ( x )}{1 + tan ^{2} ( x )} = sec (x)

p ro v e 2 sec (x) = 2 + 2 tan (x)

p ro v e sin^{2} (θ) (1 + cos^{2} (θ)) = 1

Is (4sec(x)-4)/(1-cos(θ))=4sec(x) ?
The answer to whether (4sec(x)-4)/(1-cos(θ))=4sec(x) is False