11yup...rite........Ans : A,B
if
\frac{sin^4x}{2}+ \frac{cos^4x}{2}=1/5
then
A) tan2x=2/3
B) \frac{sin^8x}{8}+ \frac{cos^8x}{127}=1/125
C) tan2x=1/3
D) \frac{sin^8x}{8}+ \frac{cos^8x}{127}=2/125
-
UP 0 DOWN 0 0 4
4 Answers
62[url=http://iitjee2009solutions.targetiit.com]Solution Key with Question Paper of 2009 IIT JEE[/url]
11sin2x = t
t2/2 + (1-t)2/3 = 1/5
On solving,
(t - 2/5)2 = 0
t = 2/5
sin2x = 2/5 ; cos2x = 3/5
tan2x = 2/3
sin8/8 + cos2x/27 = 1/125