Melody

Thanks Chris,

I am waiting for Heureka or Alan to come along and do it in 3 or 4 lines  LOL   

Prove that tan(a)/(1-cot(a))+cot(a)/(1-tan(a))= sec(A)*csc(A)+1

$$\\\frac{tan(a)}{(1-cot(a))}+\frac{cot(a)}{(1-tan(a))}= sec(A)*csc(A)+1\\\\
LHS=\frac{\frac{sina}{cosa}}{(1-\frac{cosa}{sina})}+\frac{\frac{cosa}{sina}}{(1-\frac{sina}{cosa})}\\\\
LHS=\frac{\frac{sina}{cosa}}{(\frac{sina-cosa}{sina})}+\frac{\frac{cosa}{sina}}{(\frac{cosa-sina}{cosa})}\\\\
LHS=\frac{sina}{cosa}\div (\frac{sina-cosa}{sina}) +\frac{cosa}{sina}\div(\frac{cosa-sina}{cosa}) \\\\
LHS=\frac{sina}{cosa}\times(\frac{sina}{sina-cosa}) +\frac{cosa}{sina}\times(\frac{cosa}{cosa-sina}) \\\\
LHS=\frac{sin^2a}{cosa(sina-cosa)} +\frac{cos^2a}{sina(cosa-sina)} \\\\
LHS=\frac{sin^2a}{cosa(sina-cosa)} -\frac{cos^2a}{sina(sina-cosa)} \\\\
LHS=\frac{sin^3a}{cosa*sina(sina-cosa)} -\frac{cos^3a}{cosa*sina(sina-cosa)} \\\\$$

$$\\LHS=\frac{sin^3a-cos^3a}{cosa*sina(sina-cosa)} \\\\
LHS=\frac{(sina-cosa)(sin^2a+sinacosa+cos^2a)}{cosa*sina(sina-cosa)} \\\\
LHS=\frac{(sinacosa+cos^2a+sin^2a)}{cosa*sina} \\\\
LHS=\frac{(sinacosa+1)}{cosa*sina} \\\\
LHS=1+\frac{1}{cosa*sina} \\\\
LHS=sec(a)*cosec(a)+1 \\\\
LHS=RHS\qquad QED \\\\$$

shift log 10^(-5.2) 

I just put in the input bar as 

log(10^-5.2)=

$${log}_{10}\left({{\mathtt{10}}}^{-{\mathtt{5.2}}}\right) = {\mathtt{\,-\,}}{\frac{{\mathtt{26}}}{{\mathtt{5}}}} = -{\mathtt{5.2}}$$

Excellent work anon :)

$$\left({\mathtt{4}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}\left(-{\mathtt{5}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,\times\,}}\left({\mathtt{2}}{\mathtt{\,-\,}}\left({\mathtt{3}}\right)\right){\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right){\mathtt{\,\times\,}}{\mathtt{6}}{\mathtt{\,-\,}}\left({\mathtt{2}}\right) = -{\mathtt{44}}$$

There you go "All roads lead to Rome"    

It finishes in 5.   So it is divisable by 5.  So what do you think?

anon is right but it is called atan

Arc tan is the same as inverse tan.

You can also bring it up by leaving your finger on the tan button a little longer than normal   

Do this as 2 seperate APs and then add the sums together. 

3/2 - 5k = -2(4-k)

3-10k     =  -4(4-k)

3-10k      =  -16+4k

19            =   14k

            k  =  19/14

$$\\8x^2-x^3=70\\\\
-8x^2+x^3=-70\\\\
x^3-8x^2+70=0\\\\$$