Is tan(x)-10=(sin(x)+10)/(cos(x)) ?

The answer to whether tan(x)-10=(sin(x)+10)/(cos(x)) is False

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prove tan(x)-10=(sin(x)+10)/(cos(x))

prove

tan (x) - 10 = \frac{sin ( x ) + 10}{cos ( x )}

False

Solution steps

tan (x) - 10 = \frac{sin ( x ) + 10}{cos ( x )}

Show that the two sides are not equal

For

tan (x) - 10 = \frac{sin ( x ) + 10}{cos ( x )}

plug in

x = 0

tan (0) - 10 = - 10

tan (0) - 10

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 - 10

Simplify

= - 10

\frac{sin ( 0 ) + 10}{cos ( 0 )} = 10

\frac{sin ( 0 ) + 10}{cos ( 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:

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{0 + 10}{1}

Simplify

= 10

The two sides are not equal

\Rightarrow False

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

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

1 = sec^{2} (θ) - tan^{2} (θ)

\frac{1}{sin ( θ ) cot ( θ )} = \frac{1}{cos ( θ )}

tan^{8} (x) = tan^{6} (x) sec^{2} (x) - tan^{6} (x)

Is tan(x)-10=(sin(x)+10)/(cos(x)) ?
The answer to whether tan(x)-10=(sin(x)+10)/(cos(x)) is False