1. (sin-1x+sin-1w) (sin-1y+sin-1z) = π2 then, \begin{vmatrix} X^{N_{1}} & Y^{N_{2}}\\ Z^{N_{3}} & W^{N_{4}} \end{vmatrix} where N1,N2,N3,N4 ε N :
a) has a max. value of 2
b) has a min value of -2
c) is independent of Ni
d) None
Soln & Problem: Well, It can be said that the two brackets each equal π since sin-1 x has a maximum value of π/2.
Hence, X=Y=Z=W=1 or -1
712. sin n\theta = \sum_{r=0}^{n}( b_{r} sin^{r}\theta ) for every θ , where n is odd integer, then
a) b0 = 1, b1 = 3
b) b0=0 , b1 = n
c) b0=-1 , b1 = n
d) b0 = 0, b1 = n2-3n+3
Please elaborate.
3. Prove that \sum_{r=0}^{n-1}{cosec 2^{r} \theta } = cot \frac{\theta }{2} - cot 2^{n-1} \theta
62first one is easy...
You have two cases only .. when all four x,w,y,z are +1 or -1
if all are +1, then there is only one value of the determinant.. zero..
otherwise we can have different combinations...
1 1
1 1
determinant zero
1 1
-1 1 determinant is 2
take one row as -ve you get determinant as -2
So there are 3 possiblities for the value of the determinant..
namely
0, +2, -2
71Got it what I was missing. I put the wrong values. Thanks for 1st one.
622nd one is still simpler..
just put theta = 0 you will solve half of it.. remaining half try to think more :)
623rd one is a very often repeated question...
just try to solve it for n=3...
If not then tell me will solve it :)
71Hmm.. Trying.
By the way, Are these methods the usual one to solve such types of problems or it is just an specific purpose only method.
Can you please suggest few things that we can try after looking at such sums(for solving them). I mean wherefrom to develop such tricks/idea of solving? (apart from Practice. :P LOL)