Question

If $cose{c}^6\theta -co{t}^6=aco{t}^4\theta +bco{t}^2\theta +c$ then a + b + c =

Answer 1

We know that, $a^3-b^3=(a-b)(a^2+ab+b^2)$

So, $cose{c}^6\theta -{\cot }^6\theta =(cose{c}^2\theta -{\cot }^2\theta )(cose{c}^4\theta +cose{c}^2\theta .{\cot }^2\theta +{\cot }^4\theta )=(cose{c}^4\theta +cose{c}^2\theta .{\cot }^2\theta +{\cot }^4\theta )$

Since, $cose{c}^2\theta -{\cot }^2\theta =1$

$cose{c}^4\theta +cose{c}^2\theta .{\cot }^2\theta +{\cot }^4\theta =cose{c}^2\theta (cose{c}^2\theta +{\cot }^2\theta )+{\cot }^4\theta =(1+{\cot }^2\theta )(1+2{\cot }^2\theta )+{\cot }^4\theta =1+3{\cot }^2\theta +3{\cot }^4\theta$

Hence, $a+b+c=1+3+3=7$