What is 1/((d^2+d+1)(sin(x))) ?

The solution to 1/((d^2+d+1)(sin(x))) is (csc(x))/(d^2+d+1)

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simplify 1/((d^2+d+1)(sin(x)))

simplify

\frac{1}{( d ^{2} + d + 1 ) ( sin ( x ) )}

\frac{csc ( x )}{d ^{2} + d + 1}

Solution steps

\frac{1}{( d ^{2} + d + 1 ) ( sin ( x ) )}

Use the basic trigonometric identity:

\frac{1}{sin ( x )} = csc (x)

= \frac{csc ( x )}{d ^{2} + d + 1}

(\frac{a}{2} + \frac{b}{2})^{3}

(x + y)^{2} - (a - b)^{2}

(p + 2 q)^{3} - (p - 2 q)^{3}

(a - 96)^{3}

2 x^{3} - 4 x^{2} + 2 k - 10

What is 1/((d^2+d+1)(sin(x))) ?
The solution to 1/((d^2+d+1)(sin(x))) is (csc(x))/(d^2+d+1)