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Let $f(x)=(p\sec(x))^2 + (q\csc(x))^2$

Now,$f(x)=(p\sec(x) - q\csc(x))^2+2p\sec(x)q\csc(x)$

Now, $(p\sec(x) - q\csc(x))^2\geq0$ and has minimum value = $0$ when $p\sec(x) = q\csc(x) \implies \tan(x)=\frac{q}{p} $

The minimum value of $f(x)$ will be

$2p\sec(x)q\csc(x)$ where $\tan(x)=\frac{q}{p} $

$=\frac{4pq}{\sin(2x)} =2(p^2+q^2)$

or directly putting the value $\tan(x)$, $p^2(1+\frac{q^2}{p^2})+ q^2(1+\frac{p^2}{q^2})=2(p^2+q^2)$

Again, $f(x)=(p\sec(x))^2 + (q\csc(x))^2 =p^2+q^2+ (p\tan(x))^2 + (q\cot(x))^2$

$(p\tan(x))^2 + (q\cot(x))^2 = (p\tan(x) - q\cot(x))^2+2pq $ will have minimum value $2pq$ for $\tan^2(x)=\frac{q}{p}$

Clearly, here the minimum value of $f(x)=p^2+q^2+2pq = (p+q)^2$

When I used derivative test, I reached the last value.

To rephrase the question, let h(x)=$(p\tan(x))^2 + (q\cot(x))^2$ =>f(x)=$p^2+q^2$+h(x)

As calculated above,

the minimum value of f(x)=$2(p^2+q^2)$

and the minimum value of h(x)=2pq

The minimum value of f(x) should have been $p^2+q^2+2pq$.

In general, $(p+q)^2\neq 2(p^2+q^2)$ unless $p=q$.

Where have I gone wrong?

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When you decided that you had to minimize $(p\sec(x)-q\csc(x))^2$ to minimize $f(x)$. Analogous example: $f(x)=(x)^2+(2x)$ is not minimum when $(x)^2=0$. –  Did Jul 14 '12 at 13:59
    
@did, if f(x)=g(x)+c where c is a constant, the extreme values of f(x) & g(x) will have one-one correspondence? –  lab bhattacharjee Jul 14 '12 at 14:32
    
True and unrelated since, in your case, the second term is not a constant. (Did you check the Analogous example in my comment? If not, why?) –  Did Jul 14 '12 at 14:51
    
I was talking about $(p\sec(x))^2 + (q\csc(x))^2 =p^2+q^2+ (p\tan(x))^2 + (q\cot(x))^2$. Isn't $p^2+q^2$ constant? –  lab bhattacharjee Jul 15 '12 at 4:44
1  
Zero percent accept rate - do you know about accepting answers to your questions? –  Gerry Myerson Jul 19 '12 at 7:28
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1 Answer

up vote 0 down vote accepted

Where have I gone wrong?

When you decided that you had to minimize $(p\sec(x)-q\csc(x))^2$ to minimize $f(x)$. Analogous example: $f(x)=x^2+(2x)$ is not minimum when $x^2=0$.

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