The method of eliminating `'theta'` from two given equations involving trigonometrical functions of `'theta'`. By using given equations involving `'theta'` and trigonometrical identities, we shall obtain an equation not involving `'theta'`.
On the basis of above information answer the following questions.
If `sin theta+ cos theta= a` and `sin^(3)theta+cos^(3) theta=b`, then we get `lambda a^(3) + mu b+ v a =0` when `lambda , mu, v` are independent of `theta`, then the value of `lambda^(3)+mu^(3)+v^(3) ` is

Options

A. -6
B. -18
C. -36
D. -98

Answer

To solve the problem, we need to eliminate the variable \( \theta \) from the given equations involving trigonometric functions. The equations provided are:

1. \( \sin \theta + \cos \theta = a \)
2. \( \sin^3 \theta + \cos^3 \theta = b \)

We want to express these in a form that does not involve \( \theta \) and find the values of \( \lambda \), \( \mu \), and \( v \) such that:

\[ \lambda a^3 + \mu b + v a = 0 \]

### Step 1: Square the first equation

Starting with the first equation:

\[ \sin \theta + \cos \theta = a \]

Square both sides:

\[ (\sin \theta + \cos \theta)^2 = a^2 \]

Expanding the left-hand side:

\[ \sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = a^2 \]

Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):

\[ 1 + 2 \sin \theta \cos \theta = a^2 \]

Rearranging gives us:

\[ 2 \sin \theta \cos \theta = a^2 - 1 \]

Thus, we can express \( \sin \theta \cos \theta \) as:

\[ \sin \theta \cos \theta = \frac{a^2 - 1}{2} \]

### Step 2: Use the identity for cubes

Next, we use the identity for the sum of cubes:

\[ \sin^3 \theta + \cos^3 \theta = (\sin \theta + \cos \theta)(\sin^2 \theta - \sin \theta \cos \theta + \cos^2 \theta) \]

Substituting \( \sin^2 \theta + \cos^2 \theta = 1 \):

\[ \sin^3 \theta + \cos^3 \theta = (\sin \theta + \cos \theta)(1 - \sin \theta \cos \theta) \]

Substituting \( \sin \theta + \cos \theta = a \):

\[ b = a(1 - \sin \theta \cos \theta) \]

Now substituting \( \sin \theta \cos \theta = \frac{a^2 - 1}{2} \):

\[ b = a\left(1 - \frac{a^2 - 1}{2}\right) \]

Simplifying this:

\[ b = a\left(1 - \frac{a^2}{2} + \frac{1}{2}\right) = a\left(\frac{3}{2} - \frac{a^2}{2}\right) \]

Thus:

\[ b = \frac{a(3 - a^2)}{2} \]

### Step 3: Express \( b \) in terms of \( a \)

Now we have:

\[ b = \frac{3a - a^3}{2} \]

### Step 4: Substitute into the equation

We need to express the equation:

\[ \lambda a^3 + \mu b + v a = 0 \]

Substituting \( b \):

\[ \lambda a^3 + \mu \left(\frac{3a - a^3}{2}\right) + v a = 0 \]

Multiplying through by 2 to eliminate the fraction:

\[ 2\lambda a^3 + \mu(3a - a^3) + 2va = 0 \]

Expanding gives:

\[ (2\lambda - \mu)a^3 + (3\mu + 2v)a = 0 \]

### Step 5: Compare coefficients

For this equation to hold for all \( a \), the coefficients must be zero:

1. \( 2\lambda - \mu = 0 \)
2. \( 3\mu + 2v = 0 \)

From the first equation, we can express \( \mu \) in terms of \( \lambda \):

\[ \mu = 2\lambda \]

Substituting into the second equation:

\[ 3(2\lambda) + 2v = 0 \implies 6\lambda + 2v = 0 \implies v = -3\lambda \]

### Step 6: Find \( \lambda^3 + \mu^3 + v^3 \)

Now we have:

- \( \lambda = \lambda \)
- \( \mu = 2\lambda \)
- \( v = -3\lambda \)

Calculating \( \lambda^3 + \mu^3 + v^3 \):

\[ \lambda^3 + (2\lambda)^3 + (-3\lambda)^3 = \lambda^3 + 8\lambda^3 - 27\lambda^3 = (1 + 8 - 27)\lambda^3 = -18\lambda^3 \]

Thus, the value of \( \lambda^3 + \mu^3 + v^3 \) is:

\[ \boxed{-18} \]