Question

The minimum value of (sinx+cosecx)^2 + (cosx+secx)^2

Answer 1

@yrajyalakshmi - In order to find the minimum value of

\((\sin x + \csc x)^2 + (\cos x + \sec x) ^2 \)

You should first expand the terms using \((a+b)^2 = a^2+b^2+2ab\)

\(\sin^2 x + \csc ^2 x + 2\sin x \csc x + \cos^2x + \sec^2 x +2\cos x \sec x\)
(mark this as equation 1)

Now, since, \(\sin x = \cfrac{1}{\csc x} \\
\text{OR}\\
\sin x \times \csc x = 1 \)

Similarly,

\(\cos x = \cfrac{1}{\sec x} \\
\text{OR} \\
\cos x \sec x = 1\)

Therefore, equation one becomes :

\(\sin^2 (x) + \csc^2 (x) + 2(1) + \cos^2 (x) + \sec^2 (x) + 2(1) \\
\text{Rearrange this accordingly} \\
\sin^2(x) + \cos^2(x) + \csc^2(x) + \sec^2 (x) + 4 \)

You should remember the identity that : \(\sin^2 (x) + \cos^2 (x) = 1\)

Therefore, we get :

\(1 + 4 + \csc ^2(x) + \sec^2(x) \)

\(5 + \csc^2(x) + \sec^2(x) \)

Now, I can write \(\csc^2 (x) \) as \(\cfrac{1}{\sin^2 (x)}\) and \(\sec^2(x) \) as\(\cfrac{1}{\cos^2(x)}\)

Therefore, you get :

\(5 + \cfrac{1}{\sin^2(x)} + \cfrac{1}{\cos^2(x)} \\
5 + \cfrac{\cos^2(x) + \sin^2(x) }{\cos^2(x) \times \sin^2(x)}\)

Again \(\cos^2(x) + \sin^2(x) = 1\)

Therefore, you get :

\(5 + \cfrac{1}{\cos^2(x) \times \sin^2(x)}\)

Now, we will be using sine product formula : \(\sin (2x) = 2\sin x \cos x\\ \text{OR} \\ \cfrac{1}{2} \sin(2x) = \sin x \cos x\)

Therefore,

\(5 + \cfrac{1}{\cos^2(x) \times \sin^2(x)}\)

can be written as :

\(5 + \cfrac{1}{ (\cos (x) \times \sin(x) )^2 } = 5 + \cfrac{1}{(\frac{1}{2} \sin (2x) )^2}\)

\( 5 + \cfrac{4}{(\sin(2x))^2}\)

Minimum value of sin(2x) = 1 ( in case of zero, it will be undefined , so, we will not consider the case of x = 0 .)

Put that minimum value at the place of \((\sin(2x))^2\) and thus, you will get the minimum value of the given expression.

Answer 2

While , there is a much easier method for this, called AM-GM Inequality. Have you studied it? I would be able to explain you only when you're available online.
Good Luck!