Question

The minimum value of \[ \left(1+\frac{1}{\sin^n\alpha}\right) \left(1+\frac{1}{\cos^n\alpha}\right) \] is

• A. \(1\)
• B. \(2\)
• C. \(\left(1+2^n\right)^2\)
• D. \(\left(1+2^{\frac{n}{2}}\right)^2\)

Hint:

When an expression is symmetric in \(\sin\alpha\) and \(\cos\alpha\), its minimum or maximum generally occurs at \(\sin\alpha=\cos\alpha\), provided the expression is defined.

Answer 1

The Correct Option is D

Step 1: Consider the given expression.
Let \[ E=\left(1+\frac{1}{\sin^n\alpha}\right) \left(1+\frac{1}{\cos^n\alpha}\right) \] For the expression to be defined, we must have \[ \sin\alpha\neq 0 \] and \[ \cos\alpha\neq 0 \]

Step 2: Use symmetry of the expression.
The expression contains \(\sin^n\alpha\) and \(\cos^n\alpha\) symmetrically.
So the minimum occurs when \[ \sin\alpha=\cos\alpha \] Using \[ \sin^2\alpha+\cos^2\alpha=1 \] and \[ \sin\alpha=\cos\alpha \] we get \[ 2\sin^2\alpha=1 \] Hence, \[ \sin^2\alpha=\frac{1}{2} \] So, \[ \sin\alpha=\cos\alpha=\frac{1}{\sqrt{2}} \]

Step 3: Substitute the values.
Now, \[ \sin^n\alpha=\left(\frac{1}{\sqrt{2}}\right)^n \] \[ \sin^n\alpha=2^{-\frac{n}{2}} \] Therefore, \[ \frac{1}{\sin^n\alpha}=2^{\frac{n}{2}} \] Similarly, \[ \frac{1}{\cos^n\alpha}=2^{\frac{n}{2}} \]

Step 4: Find the minimum value.
Substituting these values in \(E\), \[ E_{\min}= \left(1+2^{\frac{n}{2}}\right) \left(1+2^{\frac{n}{2}}\right) \] \[ E_{\min}= \left(1+2^{\frac{n}{2}}\right)^2 \]

Step 5: Final conclusion.
Therefore, \[ \boxed{\left(1+2^{\frac{n}{2}}\right)^2} \]