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#

We know that,
#(1)sin(90^circ-x)=sin(pi/2-x)=cosx#

#(2)cos(90^circ-x)=cos(pi/2-x)=sinx#

#(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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