trigonometry

[glyphomouni]

Here is another problem that made me land on my head:

 

if [math]sec\frac{A}{2}sec\frac{B}{2}sec\frac{C}{2}+sec\frac{A}{2}+sec\frac{B}{2}+sec\frac{C}{2} -2=0[/math] ,then one of the angles (A,B,C) of the triangle ABC is 90 degrees.

 

How do we start??

[the tree]

Well, the equation is symmetric about A, B and C for a start, so it doesn't matter which one you decide to be equal to 90. After that you'll need to look at various trig identities that you know.

[Physicsfan]

what exactly are you supposed to find?

Edited June 3, 2010 by Physicsfan

[glyphomouni]

Well, the equation is symmetric about A, B and C for a start, so it doesn't matter which one you decide to be equal to 90. After that you'll need to look at various trig identities that you know.

 

 

Thanks i will try in that direction

[the tree]

Oh I'm really sorry, I completely misread the OP. You shouldn't substitute in 90 since it's what you're trying to prove.

[Physicsfan]

Oh I'm really sorry, I completely misread the OP. You shouldn't substitute in 90 since it's what you're trying to prove.

 

now i understood the question too.

thanks:-)

now ABC is a triangle so A+B+C=180 and [imath]\frac{A+B+C}{2}=90[/imath].

[shyvera]

Here is another problem that made me land on my head:

 

if [math]sec\frac{A}{2}sec\frac{B}{2}sec\frac{C}{2}+sec\frac{A}{2}+sec\frac{B}{2}+sec\frac{C}{2} -2=0[/math] ,then one of the angles (A,B,C) of the triangle ABC is 90 degrees.

 

How do we start??

 

Have you noticed that for any x such that [math]0<x<\pi[/math] you always have [math]\sec\tfrac x2>1\,?[/math] Thus, for any angles [math]A,\,B,\,C[/math] in a triangle, we always have [math]\sec\tfrac A2\sec\tfrac B2\sec\tfrac C2+\sec\tfrac A2+\sec\tfrac B2+\sec\tfrac C2-2>2[/math] and so the left-hand side of your equation can never be equal to 0. The assertion is therefore vacuously true*.

 

*The implication [math]p\Rightarrow q[/math] is said to be vacuously true iff the statement [math]p[/math] is false.