Question

write the value of cot2 - 1/sin2

Answer 1

$[tex]3x - 5y - 4 = 0 \\ take \: x \: as \: 1 \: 2 \: \: \: 3 \: and \: 0$

1st x as 0

$3(0) - 5y - 4 = 0 \\ - 5y = 4 \\ y = \frac{4}{ - 5}$

2nd x as 1

$3(1) - 5y - 4 = 0 \\ 3 - 4 - 5y = 0 \\ - 1 - 5y = 0 \\ - 5y = 1 \\ y = \frac{1}{ - 5}$

3rd x as 2

$3(2) - 5y - 4 = 0 \\ 6 - 4 - 5y = 0 \\ 2 - 5y = 0 \\ - 5y = - 2 \\ y = \frac{2}{5}$

now last x as 3

$3(3) - 5y - 4 = 0 \\ 9 - 4 - 5y = 0 \\ 5 - 5y = 0 \\ - 5y = - 5 \\ y = \frac{5}{5} \\ y = 1$

now 2nd eqution

$2y + 7 = 9x \\ 2y + 7 - 9x = 0 \\ \\ \\ take \: x \: as \: 0 \: \: 1 \: \: 2 \: \: and \: 3$

1st x as 0

$2y + 7 - 9(0) = 0 \\ 2y + 7 = 1 \\ 2y = - 7 \\ y = \frac{ - 7}{2}$

2nd x as 1

$2y + 7 - 9(1) = 0 \\ 2y + 7 - 9 = 0 \\ 2y - 2 = 0 \\ y = 1$

3rd x as 2

$2y + 7 - 9(2) = 0 \\ 2y + 7 - 18 = 0 \\ 2y - 11 = 0 \\ 2y = 11 \\ y = \frac{11}{2}$

4th x as 3

$2y + 7 - 9(3) = 0 \\ 2y + 7 - 27 = 0 \\ 2y = 20 \\ y = \frac{20}{2} \\ y = 2$

hope you understand and like it.

mark as brainliest answer[/tex]

Answer 2

HOLA

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$cot {}^{2} \: - \: \frac{1}{sin {}^{2} } \\ \\ \\ \frac{cos {}^{2} }{sin {}^{2} } \: - \: \frac{1}{sin {}^{2} } \: = \: \frac{sin {}^{2} \: - \: cos {}^{2} }{sin {}^{2} } \\ \\ = = > \: cos {}^{2} \\ \\ so \: \: \: \: cot {}^{2} \: - \: \frac{1}{sin {}^{2} } \: = \: cos {}^{2}$

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