Algebra for IITJEE
Quadratic Equations.
(3) For relation between coefficients and zeros of polynomial ( Video Link Only )
Complex Numbers.
(1) How to find Argument of a complex number. ( Video Link Only )
(2) Modulus of complex number and properties of modulus. (Video Link Only )
(3) Conjugate of Complex Number. ( Video Link Only )
Complex Number.-Discussion Forum
$(Cos\theta +iSin\theta)^{2n+1}$ $$=\sum_{r=0}^{2n+1} {{2n+1}\choose r} Cos^{2n+1-r}\theta (i Sin\theta)^r$$ $\Rightarrow Cos(2n+1)\theta $ $$= \sum_{r=0}^{n} {{2n+1}\choose {2r} }Cos^{2n+1-2r}\theta (i Sin\theta)^{2r}$$
Permutation & Combination
From $^{(m-n)}P_r\,=\,30$ $$ $$ $(m-n)!=5.6.(m-n-r)!$ $\therefore m-n-r=4$ $$ $$ and $\, m-n=6\, ----(1)$ $$\therefore r=2$$ And $^{(m+n)}P_r\,=\,90$
Trignometry
In the equation LHS $\ge 1$ where as RHS $\le 1$ $$ $$ Hence LHS=RHS=1 is only possible solution. $\therefore x=2nπ\,,\forall n\in I$ for RHS = 1 and for same x LHS =1 if $a(2nπ)=mπ \,,m\in I$