" 20."9cos^(2)x+4sin^(2)x=5rArr tan x=

sin x + sin^(2) x + sin^(3) x = 1 rArr cos^(6) x - 4cos^(4) x + cos^(2) x =

x tan^(2)120^(@)+4cos^(2)150^(@)=9rArr x=1)3213)24)4

If cos^(2)x-2cos x-4sin x+sin2x=0 , then tan((x)/(2)) can be equal to

lim_(x rarr 0) (cos 2x - cos 7x )/(sin 3x . tan 5x) =

lim_(x rarr0)(1-cos2x+tan^(2)x)/(x sin x)

(d^(20))/(dx^(20))(2cos x cos3x) is equal to (a) 2^(20)(cos2x-2^(20)cos3x)(b)2^(20)(cos2x+2^(20)cos4x)(c)2^(20)(cos2x+2^(20)sin4x)(d)2^(20)(sin2x-2^(20)sin4x)

lim_ (n rarr0) (x sin ^ (2) 3x) / (tan ^ (2) 5x)

lim_(x rarr 0) (1-cos 5x)/(sin3x . tan 2x) =