AP ICET 2017 Shift 1 Paper

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Question : 133 of 200

Marks: +1, -0

A_x=[\table\sin x,-\cos x;\cos x,\sin x]⇒\sin x.A_x+\cos x.A_{(π/2+x)}=

$(\table 1,-1;-1,0)$
$(\table0,1;1,0)$
$(\table1,1;1,1)$
$(\table1,0;0,1)$

Solution:

Given :

A_x=[\table\sin x,-\cos x;\cos x,\sin x]

⇒A_(π/2+x)=[\table \sin(π/2+x),-\cos(π/2+x);\cos(π/2+x),\sin(π/2+x)]

⇒A_(π/2+x)=[\table \cosx,\sinx;-\sinx,\cosx]

Now,

\sin x A_{x}+\cos x A({π}/{2}+x)

=(\table \sin^2x,-\sinx\;\cosx;\sinx\;\cosx,\sin^2x)+(\table\cos^2x,\sinx\;\cosx;-\sinx\;\cosx,\cos^2x)

=(\table\sin^2x+\cos^2x,0;0,\sin^2x+\cos^2x)=(\table1,0;0,1)

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