Is cos^2(θ)(1+tan^2(θ))=cos^2(x) ?

The answer to whether cos^2(θ)(1+tan^2(θ))=cos^2(x) is False

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prove cos^2(θ)(1+tan^2(θ))=cos^2(x)

prove

cos^{2} (θ) (1 + tan^{2} (θ)) = cos^{2} (x)

False

Solution steps

cos^{2} (θ) (1 + tan^{2} (θ)) = cos^{2} (x)

Show that the two sides are not equal

For

cos^{2} (θ) (1 + tan^{2} (θ)) = cos^{2} (x)

plug in

θ = 0

For

cos^{2} (θ) (1 + tan^{2} (θ)) = cos^{2} (x)

plug in

x = 1

cos^{2} (0) (1 + tan^{2} (0)) = 1

cos^{2} (0) (1 + tan^{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

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

= 1^{2} (1 + 0^{2})

Simplify

= 1

cos^{2} (1) = 0.29192 \dots

cos^{2} (1)

Refine to a decimal form

= 0.29192 \dots

The two sides are not equal

\Rightarrow False

\frac{sin ( 4 θ ) + sin ( 2 θ )}{cos ( 4 θ ) + cos ( 2 θ )} = tan (3 θ)

\frac{sin ( x ) + tan ( x )}{sin ( x )} = 1 + sec (x)

(sec (x) + tan (x))^{2} = \frac{( csc ( x ) + 1 )}{csc ( x ) - 1}

(1 - cot^{2} (θ)) (sin^{2} (θ)) = 1 - 2 cos^{2} (θ)

\frac{sin ( x )}{- cos ( x )} = sin (x) + tan (x)

Is cos^2(θ)(1+tan^2(θ))=cos^2(x) ?
The answer to whether cos^2(θ)(1+tan^2(θ))=cos^2(x) is False