Is (sin(x))^2-(sec(x))(sin(x))^3=-1 ?

The answer to whether (sin(x))^2-(sec(x))(sin(x))^3=-1 is False

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prove (sin(x))^2-(sec(x))(sin(x))^3=-1

prove

(sin (x))^{2} - (sec (x)) (sin (x))^{3} = - 1

False

Solution steps

(sin (x))^{2} - sec (x) (sin (x))^{3} = - 1

Show that the two sides are not equal

For

sin^{2} (x) - sec (x) sin^{3} (x) = - 1

plug in

x = 0

sin^{2} (0) - sec (0) sin^{3} (0) = 0

sin^{2} (0) - sec (0) sin^{3} (0)

Use the following trivial identity:

sin (0) = 0

sin (0)

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 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

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

Simplify

= 0

The two sides are not equal

\Rightarrow False

sin (2 x) (tan (x) + cot (x)) = 2

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

\frac{cos ( θ ) + sin ( θ )}{cos ( θ )} = 1 + tan (θ)

sin (2 x) = cot (x) (1 - cos (2 x))

\frac{cos ( θ )}{1 - sin ( θ )} - tan (θ) = sec (θ)

Is (sin(x))^2-(sec(x))(sin(x))^3=-1 ?
The answer to whether (sin(x))^2-(sec(x))(sin(x))^3=-1 is False