If `(1)/(cos 290^(@))+(1)/(sqrt(3)sin 250^(@))=lambda`, then the value of `9 lambda^(4)+81 lambda^(2)+97` must be

Answer

To solve the equation \(\frac{1}{\cos 290^\circ} + \frac{1}{\sqrt{3} \sin 250^\circ} = \lambda\), we will follow these steps:

### Step 1: Rewrite the trigonometric functions
We can rewrite the trigonometric functions using their identities:
\[ \cos 290^\circ = \cos(270^\circ + 20^\circ) = \sin 20^\circ \]
\[ \sin 250^\circ = \sin(270^\circ - 20^\circ) = -\cos 20^\circ \]
Thus, we can express \(\lambda\) as:
\[ \lambda = \frac{1}{\sin 20^\circ} + \frac{1}{\sqrt{3} (-\cos 20^\circ)} \]

### Step 2: Simplify the expression for \(\lambda\)
This simplifies to:
\[ \lambda = \frac{1}{\sin 20^\circ} - \frac{1}{\sqrt{3} \cos 20^\circ} \]
To combine these fractions, we need a common denominator:
\[ \lambda = \frac{\cos 20^\circ - \frac{1}{\sqrt{3}} \sin 20^\circ}{\sin 20^\circ \cos 20^\circ} \]

### Step 3: Multiply numerator and denominator
Multiply the numerator and denominator by 2:
\[ \lambda = \frac{2(\cos 20^\circ - \frac{1}{\sqrt{3}} \sin 20^\circ)}{2 \sin 20^\circ \cos 20^\circ} \]
Using the identity \(2 \sin 20^\circ \cos 20^\circ = \sin 40^\circ\), we have:
\[ \lambda = \frac{2(\cos 20^\circ - \frac{1}{\sqrt{3}} \sin 20^\circ)}{\sin 40^\circ} \]

### Step 4: Further simplify the numerator
The numerator can be rewritten using the sine subtraction formula:
\[ \lambda = \frac{2\left(\sin 60^\circ \cos 20^\circ - \cos 60^\circ \sin 20^\circ\right)}{\sin 40^\circ} \]
This simplifies to:
\[ \lambda = \frac{2 \sin(60^\circ - 20^\circ)}{\sin 40^\circ} = \frac{2 \sin 40^\circ}{\sin 40^\circ} = 2 \]

### Step 5: Calculate \(9\lambda^4 + 81\lambda^2 + 97\)
Now substituting \(\lambda = 2\):
\[ 9\lambda^4 + 81\lambda^2 + 97 = 9(2^4) + 81(2^2) + 97 \]
Calculating each term:
\[ = 9(16) + 81(4) + 97 = 144 + 324 + 97 = 565 \]

### Final Answer
Thus, the value of \(9\lambda^4 + 81\lambda^2 + 97\) is:
\[ \boxed{565} \]