Question

Math Help ! Medals will be given !

Answer 1

@satellite73

Answer 2

the left hand side is
\[\frac{\sin(x)}{\cos(x)}+\frac{\cos(x)}{\sin(x)}\]

Answer 3

Start by using
\[\tan(x)=\frac{\sin(x)}{\cos(x)} \\ \text{and} \\ \\ \cot(x)=\frac{\cos(x)}{\sin(x)}\]
Then common denominator will get the product on denoms

Answer 4

add and you will get the answer

Answer 5

Add what ?

Answer 6

\[\frac{b}{a}+\frac{a}{b}=\frac{a^2+b^2}{ab}\]

Answer 7

add on the left

Answer 8

What's a & b

Answer 9

this is true for any \(a\) and \(b\), it is how you add, but in this example \(a=\cos(x)\) and \(b=\sin(x)\)

Answer 10

in other words, add
\[\frac{\sin(x)}{\cos(x)}+\frac{\cos(x)}{\sin(x)}\]

Answer 11

clear or not?

Answer 12

I'm still confused. Can you show me step by on how to get my answer

Answer 13

\[\frac{\sin(x)}{\cos(x)}+\frac{\cos(x)}{\sin(x)}=\frac{\sin(x)\times \sin(x)+\cos(x)\times \cos(x)}{\cos(x)\times \sin(x)}\]

Answer 14

Theres no real numbers in that problem so what wil my answer be ?

Answer 15

or
\[\frac{\sin^2(x)+\cos^2(x)}{\sin(x)\cos(x)}\]

Answer 16

what you have to do is not figure out what number this is
what you need is to "show" that the expression on the left hand side of the equal sign,
\[\tan(x)+\cot(x)\] is equal to the expression on the right hand side of the equal sign
\[\frac{1}{\sin(x)\cos(x)}\]

Answer 17

most of it is algebra
i wrote the algebra above

Answer 18

the only trig step is to rewrite
\[\tan(x)+\cot(x)\] as
\[\frac{\sin(x)}{\cos(x)}+\frac{\cos(x)}{\sin(x)}\]

Answer 19

and also to recognize the numerator when you add \(\sin^2(x)+\cos^2(x)=1\)

Answer 20

I don't have a calulator for this ?

Answer 21

@satellite73 What's my answer ?

Answer 22

@satellite73