341
Hari Shankar
·2009-11-27 05:27:05
Welcome back! We have not forgotten you either.
So we still insist you show some working before we post our replies
62
Lokesh Verma
·2009-11-27 06:28:40
I ditto prophet sir's feelings
1
champ
·2009-11-27 09:42:17
Why such double behaviour with me ??I am still in 11th class,so please have some mercy on me
62
Lokesh Verma
·2009-11-29 21:29:40
\\1=sin^2x+cos^2x=cos^2y+4/3 sin^2z = 1- sin^2y+4/3 sin^2z \\1= 1- tan^2z/6+4/3 sin^2z \\hence, tan^2z=8sin^2z \\cos^2z=1/8 \\sin^2z=7/8
Now finish it off .. [1]
1
champ
·2009-12-01 22:55:05
Continuing from where Sir left,
4sin2z=3cos2x
4*78=3cos2x
cos2x=76>1
How is this possible ?
341
Hari Shankar
·2009-12-05 02:37:53
Write the equations as
\cos^2 x = \sin^2 y...............1
6 \sin^2 y = \tan^2 z.................2 and
3 \cos^2 x = 4 \sin^2 z..............3
From 1 and 2,
6 \cos^2 x = \tan^2 z................4
Dividing 3 by 4, we get \cos^2 z = \frac{1}{8}
but of course, cos2x = 7/6 is a mistake. Have you copied the problem correctly? In any case this is the way to go about it