Question

solve this question ....

Answer 1

$\mathfrak{HELLO\:\: DEAR,}$
$\mathfrak{\boxed{5}}$
$\mathit{3(sinx - cosx)^{4} + 6(sinx + cosx)^{2} + 4(sin^{6}x + cos^{6}x)} \\ \\ \mathit{= 3 [(sin x - cos x)^2]^2 + 6[(sin^2x + cos^2x) + 2 sin x* cos x] + 4 [(sin^{2} x)^3 + (cos^{2} x)^3]} \\ \\ \mathit{=3[(sin^2x + cos^2x) - 2 sin x cos x]^2 + 6(1 + 2 sin x cos x) + 4 [(sin^2x + cos^2x)(sin^4x + cos^4x - sin^2x cos^2x)]} \\ \\ \mathit{=3(1-2 sin x cosx)^2+6+12 sin x cosx + 4 [(sin2x)^2 + (cos2x)^2 + 2sin^2x cos^2x - 3 sin^2x cos^2x]}\\ \\ \mathit{= 3 [1 + 4 sin^2x cos^2x-4 sin x cos x)] + 6 + 12 sin x cos x + 4 [(sin^2x + cos^2x) - 3sin^2xcos^2x]} \\ \\ \mathit{= 3 + 12 sin^2x cos^2x-12 sin x cos x + 6 + 12 sinx cosx + 4- 12 sin^2x cos^2x} \\ \\ \mathit{= 13}$

$\mathfrak{\boxed{6}}$

$\mathit{WE \: \: KNOW \: \: THAT:- SEC^2\theta \geqslant1 }$

$\frac{(4x)}{(x + y)^2} \geqslant 1$
$\mathit{4xy \geqslant (x + y)^2} \\ \\ \mathit{4xy \geqslant x^2 + y^2 + 2xy} \\ \\ \mathit{0 \geqslant x^2 + y^2 - 2xy} \\ \\ \mathit{0\geqslant (x - y)^2}$
$\mathit{THE \: \: SQUARE \: \: OF \: \: ANY \: \: REAL \: \: VALUE \: \: IS \: \: NON \: \: NEGATIVE,} \\ \\\mathit{THE \: \: ONLY \: \: CASE \: \: IN \: \: WHICH \: \: THIS \: \: EXPRESSION \: \: CAN \: \: BE \: \: TRUE \: \: IS \: \: THE \: \: CASE \: \: OF \: \: (x-y)^{2} = 0} \\ \\ \mathit{\therefore SEC ^{2} A = 4xy/(x + y)^{2} CAN \: \: BE \: \: TRUE \: \: ONLY \: \: FOR \: \: THE \: \: CASE \: \: (C) x = y}$

$\large{\mathbf{\underline{I \: \: HOPE \: \: ITS \: \: HELP \: \: YOU \: \: DEAR, \: \: THANKS}}}$

Answer 2

☆Hey friend!!!!!! ☆

$|5 | \\ \\ 3( {sinx - cosx)}^{4} + 6( {sinx + cosx)}^{2} + 4( {sin}^{6}x + {cos}^{6} x) \\ \\ = 3 {(( {sinx - cosx)}^{2} )}^{2} + 6(( {sin}^{2}x + {cos}^{2}x ) + 2sinx.cosx) + 4( { {sin}^{2}x) }^{3} + ( { {cos}^{2}x) }^{3} ) \\ \\ = 3(( {sin}^{2} x + {cos}^{2} x) - 2sinx.cosx)^{2} + 6(1 + 2sinx.cosx) + 4(( {sin}^{2} x + {cos}^{2} x)( {sin}^{4} x + {cos}^{4} x - {sin}^{2} x. {cos}^{2} x)) \\ \\ = 3( {(1 - 2sinx.cosx)}^{2} + 6 + 12sinx.cosx + 4(( {sin2x)}^{2} + ( {cos2x)}^{2} + 2 {sin}^{2} x. {cos}^{2} x - 3 {sin}^{2} x. {cos}^{2} x) \\ \\ = 3(1 + 4 {sin}^{2} x. {cos}^{2} x - 4sinx.cosx) + 6 + 12sinx.cosx + 4( {sin}^{2} x + {cos}^{2} x) - 3 {sin}^{2} x. {cos}^{2} x) \\ \\ = 3 + 12 {sin}^{2} x. {cos}^{2} x - 12sinx.cosx + 6 + 12sinx.cosx + 4 - 12 {sin}^{2} x. {cos}^{2} x) \\ \\ = > 3 + 6 + 4 \\ \\ = > 13 \: \: \: \: \: .....answer \\ \\ \\ \\ \\ \\ |6| \\ \\ we \: know \: that \: \\ \\ \frac{4xy}{ {(x + y)}^{2} } \geqslant 1 \\ \\ = > 4xy \geqslant {(x + y)}^{2} \\ \\ 4xy \geqslant {x}^{2} + {y}^{2} + 2xy \\ \\ 0 \geqslant {x}^{2} + {y}^{2} - 4xy + 2xy \\ \\ = > 0 \geqslant {x}^{2} + {y}^{2} - 2xy \\ \\ = > 0 \geqslant {(x - y)}^{2} \\ \\ we \: know \: = > \: the \: square \: of \: any \: real \: number \: is \: always \: positive$

this expression can be true is the case :-

sec2A = 4xy/(x+y)2

can be true only for the case " x=y"

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I think my answer is capable to clear your confusion ☺☺☺☺☺
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Devil_king ▄︻̷̿┻̿═━一