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r Cos[γ[t]] + 1/2 l Sqrt[1 - (r^2 Sin[γ[t]]^2)/l^2]

How can I ask Mathematica to write my expression without l^2 in the denominator such as:

r Cos[γ[t]] + 1/2*Sqrt[l^2 - r^2 Sin[γ[t]]^2]

Sorry, if my question is very simple, but I have been unable to work out an answer.

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closed as off-topic by m_goldberg, user9660, Bob Hanlon, Öskå, MarcoB Dec 29 '15 at 0:10

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – m_goldberg, Community, Bob Hanlon, Öskå, MarcoB
If this question can be reworded to fit the rules in the help center, please edit the question.

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r Cos[γ[t]] + 1/2 l Sqrt[1 - (r^2 Sin[γ[t]]^2)/l^2]
Simplify[%, Assumptions -> {l^2 > 0, l > 0}]

Or:

r Cos[γ[t]] + 1/2 l Sqrt[1 - (r^2 Sin[γ[t]]^2)/l^2]
Simplify[%, Assumptions -> {Re[l] > 0, Im[l] == 0}]

But I am sure this is discussed in Simplify help page.

EDIT

As @BobHanlon pointed out l>0 implies l^2>0, so one should write:

Simplify[%, Assumptions -> {l > 0}]
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  • 1
    $\begingroup$ @Bendesarts You are welcome, but you might also accept it ;) $\endgroup$ – mattiav27 Dec 28 '15 at 13:41

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