Is sin(x)tan(x)=1+cos(x) ?

The answer to whether sin(x)tan(x)=1+cos(x) is False

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prove sin(x)tan(x)=1+cos(x)

prove

sin (x) tan (x) = 1 + cos (x)

False

Solution steps

sin (x) tan (x) = 1 + cos (x)

Show that the two sides are not equal

For

sin (x) tan (x) = 1 + cos (x)

plug in

x = 0

sin (0) tan (0) = 0

sin (0) tan (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:

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

= 0 \cdot 0

Simplify

= 0

1 + cos (0) = 2

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

= 2

The two sides are not equal

\Rightarrow False

(\frac{1 + sin ( x )}{cos ( x )})^{2} + (\frac{1 - sin ( x )}{cos ( x )})^{2} = 2 + 4 tan^{2} (x)

sin (x + y) - sin (x - y) = 2 sin (y) cos (x)

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

sin (2 a) = 2 cos (a) sin (a)

sin (2 α) = \frac{2 tan ( α )}{1 + tan ^{2} ( α )}