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??

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.

what exactly are you supposed to find?

Edited June 3, 201015 yr by Physicsfan

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

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.

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].

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.