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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove sin(x)tan(x)=(1+cos^2(x))/(cos(x))

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

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

\frac{1 + cos ^{2} ( 0 )}{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

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

Simplify

= 2

The two sides are not equal

\Rightarrow False

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

cos (2 x) = cos^{4} (x) - sin^{4} (x)

csc (\frac{π}{2} - u) = sec (u)

- 2 sin (k - l) sin (k + l) = cos (2 k) - cos (2 l)

2 tan (2 x) sin^{2} (x) = tan (2 x) - sin (2 x)

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