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

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

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prove 2tan^2(x)+sec^2(x)-2=0

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

x = 0

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

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

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

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

= 2 \cdot 0^{2} + 1^{2} - 2

Simplify

= - 1

The two sides are not equal

\Rightarrow False

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tan (\frac{π}{3}) sin (\frac{2 π}{3}) = 1 + sin (\frac{π}{6})

2 cot (x) + cos (x) - sec (x) = cot (x)

(tan (θ) + cot (θ)) sin (θ) cos (θ) = 1

3 (tan (x)) + 12 = 3

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