Find the value of (sin 24° cos 6°-sin 6° sin 66°)/(sin 21° cos 39°- cos 51° sin 69°)?

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Answer 1 · 2018-06-03T19:28:08

$X = - 1$ $X=-1$

We know that,
$(1) sin (90^{\circ} - x) = sin (\frac{π}{2} - x) = cos x$ $(1)sin(90^circ-x)=sin(pi/2-x)=cosx$

$(2) cos (90^{\circ} - x) = cos (\frac{π}{2} - x) = sin x$ $(2)cos(90^circ-x)=cos(pi/2-x)=sinx$

$(3) sin A cos B - cos A sin B = sin (A - B)$ $(3)sinAcosB-cosAsinB=sin(A-B)$

We take,

$X=(sin 24° cos 6°-sin 6° sin 66°)/(sin 21° cos 39°- cos 51° sin 69°)$

$"using" : (1) and (2)$

$sin66^circ=sin(90^circ-24^circ)=cos24^circ$

$cos51^circ=cos(90^circ-39^circ)=sin39^circ$

$sin69^circ=sin(90^circ-21^circ)=cos21^circ$

So,

$X=(sin 24° cos 6°-sin 6° cos24°)/(sin 21° cos 39°- sin39° cos21°)$

$X=(sin(24^circ-6^circ))/sin(21^circ-39^circ)...to[Apply (3)]$

$X=(sin18^circ)/sin(-18)$

$X=(sin18^circ/-sin18^circ)...to[because sin(-theta)=-sintheta]$

$X=-1$

Answer 2 · 2018-06-03T20:12:02

-1

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