Question

tan5x +tan x/1-tan5xtanx

Answer 1

:)

Answer 2

just notice
\[\tan(a+b)=\frac{\tan a+\tan b}{1-tana \tan b}\]

Answer 3

a. tan 5x
b. tan x
c. tan 4x
d. tan 6x
which one is the answer? to the original problem

Answer 4

please and thank you

Answer 5

try with hint
let a=5x and b=x

Answer 6

what will u get?

Answer 7

d?

Answer 8

thats right

Answer 9

Write sin3x cos4x + cos3x sin4x in terms of a single trigonometric function.

Answer 10

hints please

Answer 11

\[\sin(a+b)=\sin a \cos b+\cos a \sin b\]

Answer 12

a. cos 7x
b. sin(-x)
c. sin 7x
d. cos(-x)

is the answer c?

Answer 13

i meant b

Answer 14

c is right

Answer 15

simplfy cos [theta + pi/3]?

Answer 16

\[\cos (a+b)=\cos a \cos b-\sin a \sin b\]

Answer 17

\[\tan 5\pi/6 +\tan \pi/6 \over 1 - \tan 5\pi/6\tan \pi/6\]

Answer 18

is it costheta + 1/2? to the simplify answer?

Answer 19

\[\cos(\theta+\frac{\pi}{3})=\frac{1}{2}\cos \theta-\frac{\sqrt{3}}{2} \sin \theta\]

Answer 20

given sin 4/5 in quadrant 1 and cos -12/13 in quadrant 2 find cos (a+b)

Answer 21

a. -16/65
b. -56/65
c. 16/65
d. 56/65

Answer 22

@mathwhiz99 @lgbasallote @Hero @KingGeorge @satellite73

One of them can help you when they get online, I'm sure :)

Answer 23

somone hurry help answer this

Answer 24

I think you meant
sin(a) = 4/5 and a is in Q1
cos(b) = -12/13 and b is in Q2

Since a is in Q1, cos(a)>0, and since b is in Q2, sin(b)>0
Therefore, sin(a) = 4/5 => cos(a) = 3/5 and cos(b) = -12/13 => sin(b) = 5/13
Cos(a + b) = cos(a)cos(b) - sin(a)sin(b) = (3/5)(-12/13) - (4/5)(5/13)
= (-36/65) - (20/65) = -56/65

The answer is b.