Stastement - 1 `alpha and beta` are tow distinct solutions of the equations `a cosx+b sin x=c, then tan ((alpha+beta)/2)` is independent for c,
Statement 2. Solution `cosx+bsinx=c` is possible, if `-sqrt(a^2+b^2)le C le sqrt(a^2+b^2)`

Options

A. (a)Both Statement I and Statement II are individually true and R is the correct explanation of Statement I.
B. (b)Both Statement I and Statement II are individually true but Statement II is not the correct explanaton of Statement I.
C. (c)Statement I is true but Statement II is false.
D. (d)Statement I is false but Statement II is true.

Answer

To solve the problem step by step, we will analyze the statements given and derive the necessary conclusions.

### Step 1: Analyze the first statement
We start with the equation given in Statement 1:
\[ a \cos x + b \sin x = c \]

### Step 2: Rewrite the equation
We can express this equation in terms of tangent:
\[ \cos x = \frac{1}{\sqrt{1 + \tan^2 x}} \]
\[ \sin x = \frac{\tan x}{\sqrt{1 + \tan^2 x}} \]

Let \( t = \tan\left(\frac{x}{2}\right) \). Using the half-angle formulas:
\[ \cos x = \frac{1 - t^2}{1 + t^2} \]
\[ \sin x = \frac{2t}{1 + t^2} \]

Substituting these into the original equation gives:
\[ a \left(\frac{1 - t^2}{1 + t^2}\right) + b \left(\frac{2t}{1 + t^2}\right) = c \]

### Step 3: Clear the denominator
Multiplying through by \( 1 + t^2 \):
\[ a(1 - t^2) + 2bt = c(1 + t^2) \]
This simplifies to:
\[ a - at^2 + 2bt = c + ct^2 \]
Rearranging gives:
\[ (c + a)t^2 + 2bt + (a - c) = 0 \]

### Step 4: Identify the roots
This is a quadratic equation in \( t \):
\[ (c + a)t^2 + 2bt + (a - c) = 0 \]
Let \( t_1 = \tan\left(\frac{\alpha}{2}\right) \) and \( t_2 = \tan\left(\frac{\beta}{2}\right) \) be the roots.

### Step 5: Use Vieta's formulas
From Vieta's formulas, we know:
- The sum of the roots \( t_1 + t_2 = -\frac{2b}{c + a} \)
- The product of the roots \( t_1 t_2 = \frac{a - c}{c + a} \)

### Step 6: Find \( \tan\left(\frac{\alpha + \beta}{2}\right) \)
Using the identity for the tangent of the sum of angles:
\[ \tan\left(\frac{\alpha + \beta}{2}\right) = \frac{t_1 + t_2}{1 - t_1 t_2} \]

Substituting the values from Vieta's:
\[ \tan\left(\frac{\alpha + \beta}{2}\right) = \frac{-\frac{2b}{c + a}}{1 - \frac{a - c}{c + a}} \]

### Step 7: Simplify the expression
Simplifying the denominator:
\[ 1 - \frac{a - c}{c + a} = \frac{(c + a) - (a - c)}{c + a} = \frac{2c}{c + a} \]

Thus:
\[ \tan\left(\frac{\alpha + \beta}{2}\right) = \frac{-\frac{2b}{c + a}}{\frac{2c}{c + a}} = -\frac{b}{c} \]

### Conclusion for Statement 1
This shows that \( \tan\left(\frac{\alpha + \beta}{2}\right) \) is independent of \( c \), validating Statement 1.

### Step 8: Analyze Statement 2
Statement 2 states that the solution \( a \cos x + b \sin x = c \) is possible if:
\[ -\sqrt{a^2 + b^2} \leq c \leq \sqrt{a^2 + b^2} \]

### Step 9: Validate Statement 2
The maximum value of \( a \cos x + b \sin x \) is given by \( \sqrt{a^2 + b^2} \) (using the Cauchy-Schwarz inequality). Therefore, for the equation to have real solutions, \( c \) must lie within the bounds stated.

### Final Conclusion
Both statements are true, but Statement 2 does not explain Statement 1. Therefore, the correct option is that both statements are true, but Statement 2 is not the correct explanation for Statement 1.

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