1Why such double behaviour with me ??I am still in 11th class,so please have some mercy on me
If sinx=cosy,√6siny=tanz,2sinz=√3cosx ; u,v,w denote respectively sin2x,sin2y,sin2z,then value of triplet (u,v,w) is ?
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7 Answers
341Welcome back! We have not forgotten you either.
So we still insist you show some working before we post our replies
1
62\\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]
1Continuing from where Sir left,
4sin2z=3cos2x
4*78=3cos2x
cos2x=76>1
How is this possible ?
341Write 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