Q1......................
for 0< \theta <\pi /2
the solutions of \sum_{m=1}^{6}{cosec(\theta +\frac{(m-1)\pi }{4}}).cosec(\theta +\frac{m\pi }{4})=4\sqrt{2}
are
1) pi/4
2)pi/6
3)pi/12
4)5pi/12
P.S i want a shorter method for this....if der is any
Q2...........................................
I_{n}=\int_{-\pi }^{\pi }{\frac{sin(nx) }{(1+\pi ^{x})sin(x)}}dx
n=0,1,2.....
then
A) I_{n}=I_{n+2}
B) \sum_{m=1}^{10}{I_{2m+1}}=10\pi
C)\sum_{m=1}^{10}{I_{2m}}=0
D)I_{n}=I_{n+1}
3411 one: you want a shorter method than what? telescopic summation is the way to go
62it means something like
(a-b)+(b-c)+(c-d)+(d-e)+....
so that every alternative one cancels out.
112)
I_n=\int_{-\pi}^{\pi}\frac{sin(nx)}{(1+\pi^x)(sinx)}dx
Now we apply properties of definite integrals to get
2I_{n}=\int_{-\pi}^{\pi}{\frac{sin(nx)}{sinx}dx}
Now write sin(nx)=sin(2x+(n-2)x).....thus (if I've not made any silly mistake)
\boxed{I_{n+2}=I_n} if n be even and option d0 if n be odd......
