Question!!
If sinA + sin²A = 1 & a cos¹²A + b cos⁸A + c cos⁶A - 1 = 0. Then find the value of the given expression.
(i) [tex]\sf b+\dfrac{c}{a}+b[/tex]
Share
Question!!
If sinA + sin²A = 1 & a cos¹²A + b cos⁸A + c cos⁶A - 1 = 0. Then find the value of the given expression.
(i) [tex]\sf b+\dfrac{c}{a}+b[/tex]
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Verified answer
[tex]\large\underline{\sf{Solution-}}[/tex]
Given that,
[tex]\sf :\longmapsto\:sinA + {sin}^{2}A = 1[/tex]
can be rewritten as
[tex]\sf :\longmapsto\:sinA = 1 - {sin}^{2}A [/tex]
We know,
[tex]\rm :\longmapsto\:\boxed{ \tt{ \: {sin}^{2}x + {cos}^{2}x = 1 \: }}[/tex]
So,
[tex]\sf :\longmapsto\:sinA = {cos}^{2}A [/tex]
Now, on squaring both sides, we get
[tex]\sf :\longmapsto\: {(sinA)}^{2} = {( {cos}^{2} A)}^{2} [/tex]
[tex]\sf :\longmapsto\: {sin}^{2}A = {cos}^{4}A[/tex]
[tex]\sf :\longmapsto\:1 - {cos}^{2}A = {cos}^{4}A[/tex]
[tex]\rm :\longmapsto\: {cos}^{4}A + {cos}^{2}A = 1[/tex]
On cubing both sides, we get
[tex]\rm :\longmapsto\: ({cos}^{4}A + {cos}^{2}A)^{3} = 1[/tex]
We know,
[tex]\boxed{ \tt{ \: {(x + y)}^{3} = {x}^{3} + {3x}^{2}y + {3xy}^{2} + {y}^{3} \: }}[/tex]
or
[tex]\boxed{ \tt{ \: {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y) \: }}[/tex]
So, using this identity, we get
[tex]\rm :\longmapsto\: {cos}^{12}A + {cos}^{6}A + 3( {cos}^{4}A)( {cos}^{2}A)( {cos}^{4}A + {cos}^{2}A) = 1[/tex]
[tex]\rm :\longmapsto\: {cos}^{12}A + {cos}^{6}A + 3( {cos}^{6}A)( 1) = 1[/tex]
[tex]\rm :\longmapsto\: {cos}^{12}A + {cos}^{6}A + 3{cos}^{6}A = 1[/tex]
[tex]\rm :\longmapsto\: {cos}^{12}A +4{cos}^{6}A = 1[/tex]
[tex]\rm :\longmapsto\: {cos}^{12}A +4{cos}^{6}A - 1 = 0[/tex]
So, on comparing with
[tex]\rm :\longmapsto\: a{cos}^{12}A + b {cos}^{8}A+c{cos}^{6}A - 1 = 0[/tex]
We get
[tex]\rm :\longmapsto\:a = 1[/tex]
[tex]\rm :\longmapsto\:b = 0[/tex]
[tex]\rm :\longmapsto\:c = 4[/tex]
Now, Consider
[tex]\rm :\longmapsto\:b+\dfrac{c}{a}+b[/tex]
On substituting the values of a, b and c, we get
[tex]\rm \: = \:0 + \dfrac{4}{1} + 0[/tex]
[tex]\rm \: = \:4[/tex]
Hence,
[tex]\rm \implies\:\boxed{ \tt{ \: b+\dfrac{c}{a}+b = 4 \: }}[/tex]