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Don't worry Mr. Carrotsticks. I just have to finish my belonging essay (I knowlol nobody has yet tried my question ='(
Wait a second... this looks like the Ruse Extension 2 2011 paper lol.Next question:
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Use the tan(A+B) thing where tanA=5 and tanB = 3, but twice. After some simplification, we have tan(A+B+C) = 0.Correct! But Ext. 1 students can do it.
Last Q of one of our mx1 papers ;
Two circles intersect at A and B. P is a point on the first circle and Q is a point on the second circle such that PAQ is a straight line. C is a point on the second circle. The tangent to the first circle at P meets QC produced to R.
Prove that PBCR is a cyclic quadrilateral
Should really justify the 3rd last line being because the exponential function is continuous.2 and e2
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For the record, to indicate inverse tan, we use arctan (x)Yahoo hoo! This one looks fun.
Hmmm...
1/2(ln(x^2+4)) + tan*-1 (x/2) + c (BTW that tan thing is meant to be inverse tan.)
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Laugh out loud.For the record, to indicate inverse tan, we use arctan (x)
ie: arctan (pi/4) = 1
wait wtf, did I just say that arctan (pi/4) = 1?Laugh out loud.
Clearly someone needs sugar or sleep.wait wtf, did I just say that arctan (pi/4) = 1?
Oh dear this is what 2 hours of sleep gets you.
I meant arctan (1) = pi/4
Oh dear me.
This only works if n is an integer. In this case, n is allowed to be any real number. Most people would fall for this trap...