1Continuing from where Sir left,
4sin2z=3cos2x
4*78=3cos2x
cos2x=76>1
How is this possible ?
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
1Why such double behaviour with me ??I am still in 11th class,so please have some mercy on me
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]
1341Write 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