What is expand-((tan(θ)-3)^2)/2 ?

The solution to expand-((tan(θ)-3)^2)/2 is -(tan^2(θ))/2+3tan(θ)-9/2

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expand-((tan(θ)-3)^2)/2

expand

- \frac{( tan ( θ ) - 3 ) ^{2}}{2}

- \frac{tan ^{2} ( θ )}{2} + 3 tan (θ) - \frac{9}{2}

Solution steps

- \frac{( tan ( θ ) - 3 ) ^{2}}{2}

(tan (θ) - 3)^{2} = tan^{2} (θ) - 6 tan (θ) + 9

= - \frac{tan ^{2} ( θ ) - 6 tan ( θ ) + 9}{2}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{tan ^{2} ( θ ) - 6 tan ( θ ) + 9}{2} = - (\frac{tan ^{2} ( θ )}{2}) - (- \frac{6 tan ( θ )}{2}) - (\frac{9}{2})

= - (\frac{tan ^{2} ( θ )}{2}) - (- \frac{6 tan ( θ )}{2}) - (\frac{9}{2})

Remove parentheses:

(a) = a, - (- a) = a

= - \frac{tan ^{2} ( θ )}{2} + \frac{6 tan ( θ )}{2} - \frac{9}{2}

Divide the numbers:

\frac{6}{2} = 3

= - \frac{tan ^{2} ( θ )}{2} + 3 tan (θ) - \frac{9}{2}

\frac{( 3 x + 5 ) ( 3 x - 5 )}{3 ( 2 x - 5 )}

\frac{4 ( x + 5 ) ( x + 1 )}{( x + 3 ) ( x - 3 )}

6 (15)

(3 x + 4 y - 2 z = 5,) - (3 x - 9 y + 8 z = 9)

1 0^{5}

What is expand-((tan(θ)-3)^2)/2 ?
The solution to expand-((tan(θ)-3)^2)/2 is -(tan^2(θ))/2+3tan(θ)-9/2