Is 3sec^2(x)-2tan^2(x)-4=0 ?

The answer to whether 3sec^2(x)-2tan^2(x)-4=0 is False

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prove 3sec^2(x)-2tan^2(x)-4=0

prove

3 sec^{2} (x) - 2 tan^{2} (x) - 4 = 0

False

Solution steps

3 sec^{2} (x) - 2 tan^{2} (x) - 4 = 0

Show that the two sides are not equal

For

3 sec^{2} (x) - 2 tan^{2} (x) - 4 = 0

plug in

x = 0

3 sec^{2} (0) - 2 tan^{2} (0) - 4 = - 1

3 sec^{2} (0) - 2 tan^{2} (0) - 4

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

Use the following trivial identity:

tan (0) = 0

tan (0)

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= 0

= 3 \cdot 1^{2} - 2 \cdot 0^{2} - 4

Simplify

= - 1

The two sides are not equal

\Rightarrow False

tan (θ) cot (θ) - 1 = 1 - sec (θ) cos (θ)

cos (2 θ) + tan (θ) sin (2 θ) = 1

\frac{cot ( 2 x )}{cos ( 2 x )} = csc (2 x)

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

\frac{sec ( t ) - 1}{1 - cos ( t )} = sec (t)

Is 3sec^2(x)-2tan^2(x)-4=0 ?
The answer to whether 3sec^2(x)-2tan^2(x)-4=0 is False