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Question

$4 s i n space 27 to the power of 0 equals$

$\sqrt{5 + \sqrt{5}} + \sqrt{3 - \sqrt{5}}$
$\sqrt{5 - \sqrt{5}} + \sqrt{3 + \sqrt{5}}$
$\sqrt{5 + \sqrt{5}} - \sqrt{3 - \sqrt{5}}$
$\sqrt{5 + \sqrt{5}} + \sqrt{3 + \sqrt{5}}$

The correct answer is: $square root of 5 plus square root of 5 end root minus square root of 3 minus square root of 5 end root$

Step by step solution:
$open parentheses sin space 27 degree plus cos space 27 degree close parentheses squared equals$ $sin squared 27 degree plus cos squared 27 degree$ $plus 2 sin 27 degree cos 27 degree$
$equals 1 plus sin 54 degree$ $open square brackets 2 sin x space cos x equals sin 2 x close square brackets$
$equals 1 plus fraction numerator square root of 5 plus 1 over denominator 4 end fraction$ $left square bracket w e space k n o w space t h a t$ $space sin 54 degree equals fraction numerator square root of 5 plus 1 over denominator 4 end fraction right square bracket$
$equals 1 fourth open parentheses square root of 5 plus 5 close parentheses$
$open parentheses sin space 27 degree plus cos space 27 degree close parentheses equals$ $equals 1 half square root of 5 plus square root of 5 end root$ -(i)
$open parentheses sin space 27 degree minus cos space 27 degree close parentheses squared equals$ $sin squared 27 degree plus cos squared 27 degree$ $negative 2 sin 27 degree cos 27 degree$
$equals 1 minus sin 54 degree$ $open square brackets 2 sin x space cos x equals sin 2 x close square brackets$
$equals 1 minus fraction numerator square root of 5 plus 1 over denominator 4 end fraction$ $left square bracket w e space k n o w space t h a t$ $space sin 54 degree equals fraction numerator square root of 5 plus 1 over denominator 4 end fraction right square bracket$
$equals 1 fourth open parentheses 3 minus square root of 5 close parentheses$
$open parentheses sin space 27 degree minus cos space 27 degree close parentheses to the power of blank equals$ $equals 1 half square root of 3 minus square root of 5 end root$ -(ii)
adding equation(i) and (ii) we get,
$2 sin space 27 degree space equals space 1 half$ [ $open parentheses square root of 5 plus square root of 5 end root close parentheses$ $plus open parentheses square root of 3 minus square root of 5 end root close parentheses$ ]
therefore, $4 space sin space 27 degree space equals$ $space open parentheses square root of 5 plus square root of 5 end root close parentheses$ $plus open parentheses square root of 3 minus square root of 5 end root close parentheses$
Hence, option(c) is the correct option.

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Consider the system of equations-x – 2y + 3z = –1–x + y – 2z = k x – 3y + 4z = 1
STATEMENT-1: The system of equations has no solution for k $not equal to$ 3
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Suppose, x > 0, y > 0, z > 0 and $capital delta$ (a, b, c) = $open vertical bar table row cell x log invisible function application 2 end cell 3 cell 15 plus log invisible function application left parenthesis a to the power of x end exponent right parenthesis end cell row cell y log invisible function application 3 end cell 5 cell 25 plus log invisible function application left parenthesis b to the power of y end exponent right parenthesis end cell row cell z log invisible function application 5 end cell 7 cell 35 plus log invisible function application left parenthesis c to the power of z end exponent right parenthesis end cell end table close vertical bar$
Statement - 1 : $capital delta$ (8, 27, 125) = 0
Statement - 2 : $capital delta$ $open parentheses fraction numerator 1 over denominator 2 end fraction comma fraction numerator 1 over denominator 3 end fraction comma fraction numerator 1 over denominator 5 end fraction close parentheses$ = 0

Suppose, x > 0, y > 0, z > 0 and $capital delta$ (a, b, c) = $open vertical bar table row cell x log invisible function application 2 end cell 3 cell 15 plus log invisible function application left parenthesis a to the power of x end exponent right parenthesis end cell row cell y log invisible function application 3 end cell 5 cell 25 plus log invisible function application left parenthesis b to the power of y end exponent right parenthesis end cell row cell z log invisible function application 5 end cell 7 cell 35 plus log invisible function application left parenthesis c to the power of z end exponent right parenthesis end cell end table close vertical bar$
Statement - 1 : $capital delta$ (8, 27, 125) = 0
Statement - 2 : $capital delta$ $open parentheses fraction numerator 1 over denominator 2 end fraction comma fraction numerator 1 over denominator 3 end fraction comma fraction numerator 1 over denominator 5 end fraction close parentheses$ = 0

parallel

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