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One way:  3 + 1/3

Another way:  10/3

I think the denominator on the LHS is sec²(x), not sec(2x), so we have:

I think I have proven that they are not equal.  Tell me if you see any errors in my woking.

Maybe my interpretation of your question was still not what you intended to ask?

$\frac{tan2x-1}{sec2x} =\frac{ tanx-cotx}{tanx+cotx }\\~\\ RHS=\left[\frac{sinx}{cosx}-\frac{cosx}{sinx} \right] \div \left[\frac{sinx}{cosx}+\frac{cosx}{sinx} \right]\\ RHS=\left[\frac{sin^2x-cos^2x}{cosxsinx} \right] \div \left[\frac{sin^2x+cos^2x}{cosxsinx}\right]\\ RHS=\left[\frac{sin^2x-cos^2x}{cosxsinx} \right] \div \left[\frac{1}{cosxsinx}\right]\\ RHS=sin^2x-cos^2x\\ RHS=-(cos^2x-sin^2x)\\ RHS=-cos(2x)\\ RHS=\frac{-1}{sec(2x)}\\ RHS \ne LHS$

LaTex:

\frac{tan2x-1}{sec2x} =\frac{ tanx-cotx}{tanx+cotx }\\~\\
RHS=\left[\frac{sinx}{cosx}-\frac{cosx}{sinx}  \right] \div \left[\frac{sinx}{cosx}+\frac{cosx}{sinx}  \right]\\
RHS=\left[\frac{sin^2x-cos^2x}{cosxsinx} \right] \div \left[\frac{sin^2x+cos^2x}{cosxsinx}\right]\\
RHS=\left[\frac{sin^2x-cos^2x}{cosxsinx} \right] \div \left[\frac{1}{cosxsinx}\right]\\
RHS=sin^2x-cos^2x\\
RHS=-(cos^2x-sin^2x)\\
RHS=-cos(2x)\\
RHS=\frac{-1}{sec(2x)}\\
RHS \ne LHS

[BXY] : [ABC] = 4 : 18

(z - 15) + 2z = 180

3z = 195

z = 65

x = 2z = 130

y = z - 15 = 50

Yea 😬

It is a tough one.   

8(3x) = 14(2x - 2)

x = 7

Thank you :) Sorry again 

It's B

Apology accepted.  Welcome to Web2.0 calc forum.   

No you didn't tell me

You said:

tan2x-1 (numerator) / sec2x (denominator) = tanx - cotx (numerator) / tanx + cotx (denominator) 

Maybe this means

A       tan2x- (1  / sec2x) = tanx -   [cotx / (tanx + cotx )]

or

B       (tan2x-1 )  / sec2x  = (tanx - cotx ) / (tanx + cotx)

Is it A or is it B ?

It is important that you learn to go between fractions and brackets.

If you are still confused then take a pic but I would like you to tell me if it is A or B anyway.  :)

Two sides of a square are consecutively based upon two straight lines $6\times x - 8\times y + 5 = 0$ and $3\times x - 4 \times y + 10 = 0$. Find the area of this square.

Hello Guest!

$6x-8y+5=0\\ f(x)=\frac{3}{4}x+0.625\\ 3x-4y+10=0\\ g(x)=\frac{3}{4}x+2.5$

$y=mx+n\\ m_{f,g}=0.75$

$\alpha =arctan(m)$

$A=a^2\\ a= (g(0)-f(0))\cdot cos(\alpha)\\ a =(g(0)-f(0))\cdot cos(arctan(m))\\ a =(2.5-0.625)\cdot cos(arctan(0.75))=\color{blue}1.5$

$A=[(2.5-0.625)\cdot cos(arctan(0.75))]^2$

$A=2.25$

  !

(a) The intersection is (2,3,-2).

(b) The answer is (3,-1,5).

The answer is -pi/2, pi/2, pi/3, 0, pi/6.

Ok first off apologies, I didn't mean to sound or be rude, I just started this website today and I didn't know I could reply to what you said so I redid the problem and posted it again. 

I told you what the denominator and numerator were. Should I take a picture of the problem? Would that be better? In the problem there are no brackets so I'm not sure were the brackets would go....

I've already tried to help you, respond on the original queston.  

Please post your repaired question here. 

Don't just ignore me and repeat the question elsewhere, that is rude.

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This is what you said in your new question:

I have to vertify the identity (there are no brackets in this problem) 

tan2x-1 (numerator) / sec2x (denominator) = tanx - cotx (numerator) / tanx + cotx (denominator) 

I still do not know what your question is, 

for example, is the numerator  tan2x-1  or     is it   -1   with the 2x seperate?   

That is what brackets are for.  Brackets would make the meaning clear.

If you have copied this from something written with fraction lines then the fraction line is used instead of brackets, if you take away the fraction line you must introduce brackets.

50^o is the supplementary angle on the other side of the 130^o angle. 

A straight line is 180^o so when you add a line that makes two angles on it, the two total 180^o.  

I feel that I'm not doing a good job of explaining it.  Use google to look up supplementary angle. 

_.

We are shifting up 3 units, so y = f(x + 3).

a_3 = 23.

My conclusion:    They would have sold more if they charged less.