Is tan^2(x)+sec^2(x)=1-cos(2x) ?

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

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prove tan^2(x)+sec^2(x)=1-cos(2x)

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

x = 0

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

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

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

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

Simplify

= 1

1 - cos (2 \cdot 0) = 0

1 - cos (2 \cdot 0)

Simplify

= 1 - cos (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

= 1 - 1

Simplify

= 0

The two sides are not equal

\Rightarrow False

cot^{2} (θ) + 1 = csc^{2} (θ)

2 cos (θ) tan (θ) csc (θ) = 2

sec (a) - sec (a) sin^{2} (a) = cos (a)

\frac{3}{sec ( x ) - tan ( x )} = 3 (sec (x) + tan (x))

\frac{( cos ^{2} ( x ) + 4 cos ( x ) + 4 )}{( cos ( x ) + 2 )} = \frac{( 2 sec ( x ) + 1 )}{sec ( x )}

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