#### Aminta_1900

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- Thread starter Aminta_1900
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18) get all three denominators to be the same.

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{cos A(1 - cot A) + sin A(1 - tan A)} / {1 - cot A - tan A + tan A*cot A}

But I don't seem to see any expressions that can be simplified from here. I am awkwardly changing the cos A to sin A*cot A, or cot to 1/tan, but getting more numerators and denominators. I can only see that the tan A*cot A at the end can be changed to +1, as tan A*(1/tan A) = (tan A)/(tan A).

Till now the questions have had mostly recognizable identities, such as 1 - cos^2 A, or sec^2 - 1, but for the first time I am seeing similar expressions to these but without the second power, but don't know how to manipulate them.

(18) Likewise with the common denominator I have

(sin^2 A + 1 + 2*cos A + cos^2 A) / (sin A + sin A*cos A)

But my thoughts again would lead me to change the sin^2 A into 1 - cos^2 A, or go all obscure with (1/sec^2 A) * (cos A*tan A). Can you go from here?

But thanks for reading thus far.

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Isn't the cotangent the reciprocal of the tangent? So shouldn't their product be simplify-able?(15) If I look for the common denominator I get

{cos A(1 - cot A) + sin A(1 - tan A)} / {1 - cot A - tan A + tan A*cot A}

Isn't there a simplification of [imath]\sin^2(x) + \cos^2(x)[/imath]?(18) Likewise with the common denominator I have

(sin^2 A + 1 + 2*cos A + cos^2 A) / (sin A + sin A*cos A)