IIT JEE 2009 P1 Q31

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

4 Answers

62
Lokesh Verma ·

[url=http://iitjee2009solutions.targetiit.com]Solution Key with Question Paper of 2009 IIT JEE[/url]

1
abhishektewari ·

the ans shoul be both A and B ...pls check it bhaiya

11
rkrish ·

yup...rite........Ans : A,B

11
rkrish ·

sin2x = 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

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