MATH math math.....only math

tan(A/2)=\frac{sin(A/2)}{cos(A/2)} \\ \text{Now consider} \frac{sin(A/2)}{cos(A/2)}(1+secA)(1+sec2A)....(1+sec2^{n}A) \\ = \frac{sin(A/2)}{cos(A/2)}\frac{(1+cosA)(1+cos(2A))(1+cos(4A)...(1+cos(2^{n}A)}{cosAcos2Acos4A....cos(2^{n}A)} \\ 1+cosA=2cos^{2}(A/2),1+cos2A=2cos^{2}(A/2),........[\text{We can now cancel the terms accordingly}] \\ \text{After simplification we get} \\ \frac{sin(2^{n}A)}{cos(2^{n}A)} \Longrightarrow tan(2^{n}A) \\ \text{Now its easy to substitute the options} \\ \text{Answers are a,b,c,d}

Please tell whether i'm correct or not!!!!

i just said that seeing 0/0 form.

yo dicthum............if u use lhospital from wher will u get awkward numbers like 11e/24or e/24 eh???

k another interesting wala......

Q 10

ya abhirup.......rite...try out 7th......

Q8.....is it 300

9*9*3 + 9*3*2 + 3= 300

Q.8 A very very very easy question to end d day with...........

Prajith..........ive specified...MULTIPLE ANSWER WALA...........and ya B is also correct....show me d working....i was confused when i tried....hehe

for ques six

just take out x² common from the underoot below
in numerator we get

x-1/x³

now put the thing inside the denominator root =t

you get the answer

Q 2.

WOW...............MS i dint think of tht.........u r gr8 da...

Q5

∫dx/√x(1+√x)(1-√x)
put x = cos²Î¸
ques is done!!!!!!!!!!!!!!!!

question six was already discussed in this site

put x= sinθ and proceed

q 4 is quite easy

ans will be A

as just the point will be exterior to circle and angle <pi/4

Q. 5 this is nice... try i dint get d answer hehe

q3 iit KA HAI
ISKA ANS (A) AND (B) HOGA
BAHUT BEHAS HUI THI ISKE UPPER

yup again correct..........its 3 twas easy naa....

ya satan........thts correct but its 3p/2 not 3/2p hehe.......

for second one
apply newton leibnitz rule

we get

-cosxsin²xf(sinx)=-cosx

thus f(sinx)=1/sin²x

f(x)=1/x²

here ans=3

by symmetry we get the probability of ocurence of a,b, and c equal =x

probability that exactly one of A or B occurs =

x(1-x) +(1-x)(x)

=2x-2x²=p

probability that all occur simultaneously = x³=p²

now the probability that exactly one of these occur =

1-(probability none occurs)

=1-(1-x)³=

1-1+x³+3x(1-x)
=x³+3x(1-x)

=p² +3/2p