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Prove that: (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0
L.H.S.
= (sin 3x + sin x) sin x + (cos 3x – cos x) cos x
= RH.S.
Prove that: 
numerator=(sin7x+sin5x)+(sin9x+sin3x)
=2sin(12x/2)cos(2x/2)+2sin(12x/2)cos(6x/2) (sinA+sinB=2sin(A+B)/2 cos(A-B)/2 )
=2sin6x cosx+2sin6x cos3x
=2sin6x(cosx+cos3x)
=2sin6x(2cos(4x/2)cos(2x/2)) (cosA+cosB=2cos(A+B)/2 cos(A-B)/2)
=4sin6x cos2x cosx
denominator=(cos7x + cos5x)+(cos9x+cos3x) (cosA+cosB=2cos(A+B)/2 cos(A-B)/2)
=2cos(12x/2)cos(2x/2)+2cOs(12X/2)Cos(6X/2)
=2cos6x cosx +2cos6x cos3x
=2cos6x(cosx+cos3x)
=2cos6x * 2cos(4x/2)cos(2x/2)
=4cos6x cos2x cosx
LHS=(4sin6x cos2x cosx)/(4cos6x cos2x cosx)
=sin6x/cos6x
=tan6x
, x in quadrant II
Here, x is in quadrant II.
i.e.,
Therefore, 
are all positive.
As x is in quadrant II, cosx is negative.
∴
Thus, the respective values of are
.
Find
for 
, x in quadrant III
Here, x is in quadrant III.
Therefore, 
and 
are negative, whereas
is positive.
Now, 
Thus, the respective values of 
are
.
Find
for 
, x in quadrant II
Here, x is in quadrant II.
Therefore,
, and 
are all positive.
[cosx is negative in quadrant II]
Thus, the respective values of
are 
.
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