think logically and not as formulas or anything. And also what you can do is to try and solve more of those if u arent comfortable with them :) .

 

If u still dont like the chapter, then u could always try a few substitutions in the sums which will easily solve them

 

P.S: My personal fav is combinatorics :D 

no i will only lead solns but not post them :D let the people who posted try?

And i will say it again we shd let others try too mr. computer001 so i think i will give leading ideas only

easy.

Hint: Multiply nr and dr by (cosecx)^2 . Nr becomes (cosecx)^2 and dr becomes sqrt(cosalpha + cotx sinalpha)

so put the dr = t and make the nr dt :)

 

well a combinatorial argument would go as follows

consider n identical elements and r-1 sticks to separate them so that they form a total of r groups with n+r-1 items.

The total number of ways of placing these r sticks is (n+r-1)C(r-1) :)

actually the answer shd be d i believe bcos the inequality goes like for a triangle,

$ (1+ \frac{b-c}{a})^a \cdot (1+ \frac{c-a}{b})^b \cdot (1+ \frac{a-b}{c})^c leq 1$

 

proof:

AM $geq$ GM

$implies$ $((1+ \frac{b-c}{a}) + (1+ \frac{b-c}{a}) +.........+ a times) + $((1+ \frac{c-a}{b}) + (1+ \frac{c-a}{b}) +.........+ b times) ...............................

----------------------------------------------------------------------------

     (a+b+c)

$geq$ the GM But we get AM = 1

 

Hence its equilateral and answer is D