Beware of blind algebra!!

Here is an interesting case of a problem in trigonometry...... I had encountered this in my CET entrance exam .....

The problem asks us to simplify. a[bcos(C)-c.cos(B)]. ......................(1).

Where a,b,c are sides of any triangle and A,B,C are its angles.

Using a= bcos(C)+c.cos(B). (Projection rule)

The expression (1) simplifies to

=b^2cos^2(C)-c^2cos^2(B)....... (2)

=b^2[1-sin^2(C)]-c^2[1-sin^2(B)]

=b^2-c^2-b^2sin^2(C)+c^2sin^2(B)............(3)

But sine rule says .....

bsin(C)=csin(B)

Implying that

b^2sin^2(C)=c^2sin^2(B)

Hence the expression (3) simplifies to

=b^2-c^2........ (4)

So far so good but comparing (4) with (2)

We may deduce cos^2(C)=1=cos^2(B)......

Implying C=B=90 degrees !!!!! .....( not a triangle!!)

Are you kidding ALGEBRA !!!!

Can you resolve the paradox????

Easy Math Editor

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Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$