Question

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Answer 1

Given

→ cos∅ + cos²∅ = 1

→ cos∅ + 1 - sin²∅ = 1

One cancels out both sides,

→ cos∅ = sin²∅

Solution

$\Longrightarrow{\mathsf{{sin}^{12}\theta + 3 {sin}^{10}\theta + 3{sin}^{8}\theta + {sin}^{6}\theta + 2 {sin}^{4}\theta + 2 {sin}^{2}\theta - 2}} \\ \\ \Longrightarrow{\mathsf{{( {sin}^{4}\theta) }^{3} + 3 { ({sin}^{4}\theta) }^{2} {sin}^{2}\theta + 3 {sin}^{4}\theta { ({sin}^{2}\theta) }^{2} + {( {sin}^{2} \theta)}^{3} + 2(( { {sin}^{2}\theta )}^{2} + {sin}^{2}\theta) - 2}} \\ \\ \Longrightarrow{\mathsf{{( {sin}^{4}\theta + {sin}^{2} \theta)}^{3} + 2( {cos}^{2}\theta + cos\theta) - 2}} \\ \\ \Longrightarrow{\mathsf{{( {cos}^{2}\theta + cos\theta)}^{3} + 2(1) - 2}} \\ \\ \Longrightarrow{\mathsf{1 + 2 - 2}} \\ \\ \Longrightarrow{\mathsf{1}}$

Hence Proved

Answer 2

Answer:

: Solution :

=sin^12∅ + 3sin∅ + 3sin^8∅ + sin^6∅+ 2sin^4∅ + 2 sin^2∅ - 2.

=(sin^4∅)^3 + 3(sin ^4∅)^2 sin^2∅ + 3sin ^4∅ ( sin ^2∅)^2 + (sin^2∅ )^3 + 2(sin^2∅ )^2 + sin^2∅ ) - 2 .

= (sin^4∅ + sin^2∅ )^3 + 2(cos^2∅ + cos∅) -2

= (cos^2∅ + cos∅)^3 + 2(1) - 2

= 1 + 2 - 2

= 1

= So it is proved .