Question

$question \: in \: attachment$
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Answer 1

Answer:

Step-by-step explanation:

i. $cot^{2} \alpha$

ii.

$x^{2} + x - 12 \\= x^{2} + 4x - 3x -12\\=x(x+4) - 3(x+4)\\=(x-3)(x+4)$

⇔$x=3$ or $x= -4$

sum of roots = -1

iii. cos∅

iv. 2

v. option a

Answer 2

Answer:

i. cot^{2} \alphacot

2

α

ii.

\begin{gathered}x^{2} + x - 12 \\= x^{2} + 4x - 3x -12\\=x(x+4) - 3(x+4)\\=(x-3)(x+4)\\\\\end{gathered}

x

2

+x−12

=x

2

+4x−3x−12

=x(x+4)−3(x+4)

=(x−3)(x+4)

⇔x=3x=3 or x= -4x=−4

sum of roots = -1

iii. cos∅

iv. 2

v. option a

ur there didi

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