If #Sin17^@ =x/y# , then show that #Sec17^@ - sin73^@ = x^2/(ysqrt((y^2 - x^2))# ?

If #Sin17^@ =x/y# , then show that #Sec17^@ - sin73^@ = x^2/(ysqrt((y^2 - x^2))#

If #Sin17^@ =x/y# , then show that #Sec17^@ - sin73^@ = x^2/(ysqrt((y^2 - x^2))#

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Dec 11, 2017

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See full explanation for solution.

Explanation:

By the cofunction identity:
#sin(73^circ)=cos(90^circ-73^circ)=cos(17^circ)#

So we have #sec(17^circ)-cos(17^circ)#. We know that #sec(x)=1/cos(x)# so we can rewrite this expression:

#1/cos(17^circ)-cos(17^circ)=(1-cos^2(17^circ))/cos(17^circ)#

By the Pythagorean ID, #sin^2(x)+cos^2(x)=1# and so #1-cos^2(x)=sin^2(x)#:

#sin^2(17^circ)/cos(17^circ)#

Rearranging #sin^2(x)+cos^2(x)=1# we can also see that #cos(x)=sqrt(1-sin^2(x))#.

(Since #17^circ# is in QI all trig functions are positive.)

#sin^2(17^circ)/sqrt(1-sin^2(17^circ))#.

Now we substitute #x/y# for #sin(17^circ)#:

#(x/y)^2/sqrt(1-(x/y)^2)#.

Now we do some algebra:

#(x^2/y^2)/sqrt(1-x^2/y^2)=(x^2/y^2)/sqrt((y^2-x^2)/y^2)#

#=(x^2/y^2)/((1/y)sqrt(y^2-x^2))=(y*(x^2/y^2))/sqrt(y^2-x^2)#

#=(x^2/y)/sqrt(y^2-x^2)=x^2/(y*sqrt(y^2-x^2))#, as required.

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dk_ch Share
Dec 11, 2017

#LHS=sec17^@-sin73^@#

#=1/(cos17^@)-sin(90^@-17^@)#

#=1/(cos17^@)-cos17^@#

#=(1-cos^2 17^@)/cos17^@#

#=(sin^2 17^@)/sqrt(1-sin^2 17^@)#

#=(x^2/y^2)/sqrt(1-x^2/y^2)#

#=x^2/y^2xxy/sqrt(y^2-x^2)#

#=x^2/(ysqrt(y^2-x^2))=RHS#

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