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secondary 4 | A Maths
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secondary 4 chevron_right A Maths chevron_right Singapore

Can some one pls help, thank

Date Posted: 1 year ago
Views: 187
Eric Nicholas K
Eric Nicholas K
1 year ago
"Use the result in part (i)..."

We start from the left side.

Make the numerator in part ii become (sin A + cos A)².

Convert cos 2A into cos² A - sin² A before factorising it into (cos A - sin A) (cos A + sin A).

Then the LHS becomes (cos A + sin A) / (cos A - sin A).

Divide both the numerator and the denominator each by cos A.

You will obtain the right side expression.
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1 year ago
Thank a lot
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1 year ago
Can further help, still cannot solve
Eric Nicholas K
Eric Nicholas K
1 year ago
cos 2A has three forms.

The cosine-sine version is useful because of the factorisation of the difference of two squares algebraic identity.

The cosine-only version is useful in the situation of 1 + cos 2A where the “1” will vanish later on.

The sine-only version is useful in the situation of 1 - cos 2A where the “1” will vanish later on.
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1 year ago
Thank for sharing

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Eric Nicholas K
Eric Nicholas K's answer
5997 answers (Tutor Details)
1st
The step in which I introduced “1/cos A” in both the numerator and denominator (equivalent to dividing by cos A on both the numerator and denominator) is vital in obtaining the final format.
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1 year ago
Thank a lot