Question

Question!!
If sinA + sin²A = 1 & a cos¹²A + b cos⁸A + c cos⁶A - 1 = 0. Then find the value of the given expression.
(i) $\sf b+\dfrac{c}{a}+b$

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\sf :\longmapsto\:sinA + {sin}^{2}A = 1$

can be rewritten as

$\sf :\longmapsto\:sinA = 1 - {sin}^{2}A$

We know,

$\rm :\longmapsto\:\boxed{ \tt{ \: {sin}^{2}x + {cos}^{2}x = 1 \: }}$

So,

$\sf :\longmapsto\:sinA = {cos}^{2}A$

Now, on squaring both sides, we get

$\sf :\longmapsto\: {(sinA)}^{2} = {( {cos}^{2} A)}^{2}$

$\sf :\longmapsto\: {sin}^{2}A = {cos}^{4}A$

$\sf :\longmapsto\:1 - {cos}^{2}A = {cos}^{4}A$

$\rm :\longmapsto\: {cos}^{4}A + {cos}^{2}A = 1$

On cubing both sides, we get

$\rm :\longmapsto\: ({cos}^{4}A + {cos}^{2}A)^{3} = 1$

We know,

$\boxed{ \tt{ \: {(x + y)}^{3} = {x}^{3} + {3x}^{2}y + {3xy}^{2} + {y}^{3} \: }}$

or

$\boxed{ \tt{ \: {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y) \: }}$

So, using this identity, we get

$\rm :\longmapsto\: {cos}^{12}A + {cos}^{6}A + 3( {cos}^{4}A)( {cos}^{2}A)( {cos}^{4}A + {cos}^{2}A) = 1$

$\rm :\longmapsto\: {cos}^{12}A + {cos}^{6}A + 3( {cos}^{6}A)( 1) = 1$

$\rm :\longmapsto\: {cos}^{12}A + {cos}^{6}A + 3{cos}^{6}A = 1$

$\rm :\longmapsto\: {cos}^{12}A +4{cos}^{6}A = 1$

$\rm :\longmapsto\: {cos}^{12}A +4{cos}^{6}A - 1 = 0$

So, on comparing with

$\rm :\longmapsto\: a{cos}^{12}A + b {cos}^{8}A+c{cos}^{6}A - 1 = 0$

We get

$\rm :\longmapsto\:a = 1$

$\rm :\longmapsto\:b = 0$

$\rm :\longmapsto\:c = 4$

Now, Consider

$\rm :\longmapsto\:b+\dfrac{c}{a}+b$

On substituting the values of a, b and c, we get

$\rm \:  =  \:0 + \dfrac{4}{1} + 0$

$\rm \:  =  \:4$

Hence,

$\rm \implies\:\boxed{ \tt{ \: b+\dfrac{c}{a}+b = 4 \: }}$