If #sin theta +sin^2theta=1# and #acos^12theta+bcos^10theta+c cos^8theta+dcos^6theta-1=0#, then #(b+c)/(a+d)=#?

Question

#a)1#
#b)2#
#c)-2#
#d)3#

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Answer 1 · 2018-01-20T23:46:44

Given

#sin theta +sin^2theta=1#

#=>sin theta =1-sin^2theta#

#=>sin theta =cos^2theta#

Again

#acos^12theta+bcos^10theta+c cos^8theta+dcos^6theta-1=0#

#=>a(cos^2theta)^6+b(cos^2theta)^5+c (cos^^2theta)^4+d(cos^2theta)^3=1^3#

#=>a(sintheta)^6+b(sintheta)^5+c (sintheta)^4+d(sintheta)^3=(sin^2theta+sintheta)^3#

#=>asin^6theta+bsin^5theta+c sin4theta+dsin^3theta=sin^6theta+3sin^5theta+3sin^4theta+sin^3theta#

Comparing the coefficients of #sin^6theta,sin^5theta,sin^4theta and sin^3theta# in both sides we get the following possible values for #sintheta!=0#

#a=1,b=3,c=3andd=1#

then

#(b+c)/(a+d)=(3+3)/(1+1)=3#

So option (d) accepted