Is csc^2(sin^2(x)-1)= 1/(cos^2(x)) ?

The answer to whether csc^2(sin^2(x)-1)= 1/(cos^2(x)) is False

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prove csc^2(sin^2(x)-1)= 1/(cos^2(x))

prove

csc^{2} (sin^{2} (x) - 1) = \frac{1}{cos ^{2} ( x )}

False

Solution steps

csc^{2} (sin^{2} (x) - 1) = \frac{1}{cos ^{2} ( x )}

Show that the two sides are not equal

For

csc^{2} (sin^{2} (x) - 1) = \frac{1}{cos ^{2} ( x )}

plug in

x = 0

csc^{2} (sin^{2} (0) - 1) = 1.41228 \dots

csc^{2} (sin^{2} (0) - 1)

Refine to a decimal form

= 1.41228 \dots

\frac{1}{cos ^{2} ( 0 )} = 1

\frac{1}{cos ^{2} ( 0 )}

Use the following trivial identity:

cos (0) = 1

cos (0)

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

= 1

= \frac{1}{1 ^{2}}

Simplify

= 1

The two sides are not equal

\Rightarrow False

3 tan^{2} (y) (csc^{2} (y) - 1) = 3

tan (θ^{2}) + 6 = sec (θ^{2}) + 5

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

1 - cot^{2} (x) sin^{2} (x) = sin^{2} (x)

csc (ω) - sin (ω) = cot (ω) cos (ω)

Is csc^2(sin^2(x)-1)= 1/(cos^2(x)) ?
The answer to whether csc^2(sin^2(x)-1)= 1/(cos^2(x)) is False