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Learn Miscellaneous Exercise 3 with Free Lessons & Tips

Prove that:

L.H.S.

= 0 = R.H.S

Comments

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.

Comments

Prove that:

L.H.S. =

Comments

Prove that:

L.H.S. =

Comments

Prove that:

It is known that.

∴L.H.S. =

Comments

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

Comments

Prove that:

L.H.S. =

Comments

, 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.

Comments

Find for , x in quadrant III

 

Here, x is in quadrant III.

Therefore, and are negative, whereasis positive.

Now,

Thus, the respective values of are.

Comments

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 .

Comments

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