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Complementary Trig ratios

A really handy thing to know here is that the co in the front of half of the trig ratios stands for complement

Complementary angles add up to 90 degrees.

In any right angled triangle there are 2 acute angles.  

If one is  θ    the other is the complement of  θ       That is (90-θ)

so

$$\\\mathbf{co}sine(\theta)=sin(90-\theta)\\
\mathbf{co}tangent(\theta)=tangent(90-\theta)\\
\mathbf{co}secant(\theta)=secant(90-\theta)\\$$

$$\\\mathbf{co}s(\theta)=sin(90-\theta)\\
\mathbf{co}t(\theta)=tan(90-\theta)\\
\mathbf{co}sec(\theta)=sec(90-\theta)\\$$

Oops, I didn't notice you'd changed to log base 10 in the question, making it different to what was in the subject.

Are these supposed to both be logs to base 10, or only one? It doesn't change the method, anyway.

Actually, Nauseated, these are known as equivalent, but indeterminate, biblical forms

every city or house divided against itself  = 0

a house divided against a house  = 0

If you set them equal to one another......the result is recurring CDD......

[ Which is a necessary quality for zealous religious belief anyway, IMHO ]

To echo what the inimitable George Carlin once said, "I leave symbols to the symbol-minded."

Yes, draw the right angled triangle and you will see. 

I think you'll be able to work it out for yourself once I show you that the equation that gave rise to what you wrote is this:

$${{\mathtt{e}}}^{{\mathtt{44}}} = {\left({\frac{{\mathtt{E}}}{{\mathtt{2}}}}\right)}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{\mathtt{E}}$$

Take logs of both sides, and you arrive at what you began the question with. So all you have to do is solve for E in the equation I wrote.

Easy enough?

Any proper number multiplied by zero is zero

Definitely not the Cat Woman becoz Cat Woman isn't that bad........isn't it?

Why is arccos (2.3) invalid?

Well  here is a rather pale right angled triangle.

AT is the hyotenuse so it is the longest side

Cos A = AB/AT = little side/longest side

How can you divide a little nummber by a big number and get an answer more than 1?    You can't!!

That is why  cos of an angel must NOT be more than 1  

and that is why arccos (2.3 ) is invalid!

17, becoz it's my birthday on 17September!

I do not understand your question.....

Your biblical quote is wrong.

Matthew 12:25

every city or house divided against itself shall not stand:

* Luke 11:17

and a house divided against a house falleth.

See, you misquoted this verse; it’s

 “. . . house divided against itself . . .”

Not

“. . . house divided by itself . . .”

The Greek word is “epi”

$$\xi \pi \iota$$

and while that can be translated as “by” for some uses, not one formal English translation uses “by” for these verses.

These quotes are from the King James version. This version was good enough for Jesus so it should be good enough for anyone else.

So, while you didn’t give a wrong math answer and the children won’t fail, What you did was worse: you misquoted scripture. Now, the little brats might go to h**l.

CDD in spiritual and philosophical form has eternal ramifications, you know.

You should repent of this grievous error!

3.14 is an approximation for pi

Interpreting this as [2xlog2] / [xlog3]=log2   multiply both sides by x log3 and divide both sides by log2

[ 2x log 2] / [log 2 ] = x log 3    simplify

2x  = x log 3     subtract log 3 from both sides

2x  - x log 3 = 0   factor

x (2 - log 3 ) = 0   so x = 0

But x cannot = 0  because it makes the denominator on the LHS of the original problem undefined....so this has no solution.....

what is 2 1/8- 1 2/7 = 47/56

2 1/8- 1 2/7 =

17 / 8  -  9 / 7      getting a common denominator of 56, we have

119/56 - 72 / 56 =

47 / 56

x^2+y^2-8y-33=0   add 33 to both sides 

x^2 + y^2 - 8y  = 33   complete the square on y

x^2 + y^2 - 8y + 16  = 33 + 16    factor

(x - 0)^2 + (y - 4)^2   = 49

The center of this circle is at (0 , 4) and the radius  = 7

csc^2x/cotx=csc(x)sec(x)    working with the LHS

1/sin^x / (cosx /sin x) =

1/ sin^2x (sinx / cos x) =

 (1 / sin x) (1 / cos x) =

csc x secx       .....and the LHS = RHS

(66+2/3)/100 of x = 22

(66+2/3)/100 * x = 22

 x = 22 * (100/(66+2/3))

$${\mathtt{22}}{\mathtt{\,\times\,}}\left({\frac{{\mathtt{100}}}{\left({\mathtt{66}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)}}\right) = {\mathtt{33}}$$

T2-T1 = T3-T2=t4-T3 =..... = d

Not sure what you mean

$$6\times 10^{-3}=6\times \frac{1}{10^3} = \frac{6}{10^3}$$

Maybe that will help ?

just a little mistake Sasini

280-66=214

$${\mathtt{56}}{\mathtt{\,-\,}}{\mathtt{8}}{\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{4}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{2}}}{{\mathtt{1}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{3}}}{{\mathtt{5}}}}\right) = {\frac{{\mathtt{214}}}{{\mathtt{5}}}} = {\mathtt{42.8}}$$

YOU HAVE TO FLIP THE SECOND ONE UPSIDE DOWN !

$$\frac{3}{7} \div \frac{5}{7} = \frac{3}{7}\times \frac{7}{5} = \frac{3}{5}$$

$$\displaystyle\sum_{n=1}^{5}\;\; \frac{(-1)^{n+1}\;n}{3^n}$$

Maybe this will help a little