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

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

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prove (tan^2(x))/(sec(x)+1)=sec(x)

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

x = 0

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

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

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

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

Simplify

= 0

sec (0) = 1

sec (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

= 1

The two sides are not equal

\Rightarrow False

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

tan^{3} (x) = (tan (x))^{3}

(csc (x) + tan (x)) cos (x) = sin (x) cot (x)

3 tan^{2} (x) + 13 tan (x) = 4

csc (\frac{π}{2} - θ) = sec (θ)

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