If `theta=3alpha` and `sin theta=(a)/(sqrt(a^(2)+b^(2))` , the value of the expression `a co sec alpha-b sec alpha` is

Options

A. (a) `(a)/(sqrt(a^(2)+b^(2))`
B. (b) `2sqrt(a^(2)+b^(2))`
C. (c) 'a+b'
D. (d) none of these.

Explanation

To solve the problem step by step, we need to find the value of the expression \( a \csc \alpha - b \sec \alpha \) given that \( \theta = 3\alpha \) and \( \sin \theta = \frac{a}{\sqrt{a^2 + b^2}} \).

### Step 1: Substitute for \(\theta\)
Since \(\theta = 3\alpha\), we can write:
\[ \sin(3\alpha) = \frac{a}{\sqrt{a^2 + b^2}} \]

### Step 2: Use the sine triple angle formula
The sine triple angle formula states:
\[ \sin(3\alpha) = 3\sin(\alpha) - 4\sin^3(\alpha) \]
Thus, we have:
\[ 3\sin(\alpha) - 4\sin^3(\alpha) = \frac{a}{\sqrt{a^2 + b^2}} \]

### Step 3: Find \(\csc \alpha\) and \(\sec \alpha\)
We know:
\[ \csc \alpha = \frac{1}{\sin \alpha} \quad \text{and} \quad \sec \alpha = \frac{1}{\cos \alpha} \]
We need to find \( a \csc \alpha - b \sec \alpha \).

### Step 4: Express \(\csc \alpha\) and \(\sec \alpha\) in terms of \(\sin \alpha\) and \(\cos \alpha\)
Substituting these into our expression gives:
\[ a \csc \alpha - b \sec \alpha = \frac{a}{\sin \alpha} - \frac{b}{\cos \alpha} \]

### Step 5: Find a common denominator
The common denominator for the two fractions is \(\sin \alpha \cos \alpha\):
\[ = \frac{a \cos \alpha - b \sin \alpha}{\sin \alpha \cos \alpha} \]

### Step 6: Substitute \(\sin(3\alpha)\) and \(\cos(3\alpha)\)
Using the identity \( \sin^2 \alpha + \cos^2 \alpha = 1 \), we can find \(\cos(3\alpha)\):
\[ \cos(3\alpha) = \sqrt{1 - \sin^2(3\alpha)} = \sqrt{1 - \left(\frac{a}{\sqrt{a^2 + b^2}}\right)^2} \]
This simplifies to:
\[ \cos(3\alpha) = \sqrt{\frac{b^2}{a^2 + b^2}} = \frac{b}{\sqrt{a^2 + b^2}} \]

### Step 7: Substitute back into the expression
Now substituting back, we have:
\[ = \frac{a \cos \alpha - b \sin \alpha}{\sin \alpha \cos \alpha} \]
We can express \(\sin(3\alpha)\) and \(\cos(3\alpha)\) in terms of \(\sin \alpha\) and \(\cos \alpha\).

### Step 8: Final simplification
Using the sine difference identity:
\[ \sin(3\alpha - \alpha) = \sin(2\alpha) = 2 \sin \alpha \cos \alpha \]
Thus, the expression simplifies to:
\[ = \frac{2\sqrt{a^2 + b^2}}{2} = \sqrt{a^2 + b^2} \]

### Final Answer
The value of the expression \( a \csc \alpha - b \sec \alpha \) is:
\[ \sqrt{a^2 + b^2} \]