If `(cos theta)/(a)=(sintheta)/(b), then (a)/(sec 2 theta)+(b)/(cosec 2 theta)` is equal to

Options

A. a
B. b
C. `a/b`
D. `a+b`

Answer

To solve the given problem step by step, we start with the equation provided in the question:

**Given:**
\[ \frac{\cos \theta}{a} = \frac{\sin \theta}{b} \]

### Step 1: Cross-multiply to find a relationship between \(a\) and \(b\).
Cross-multiplying gives:
\[ b \cos \theta = a \sin \theta \]

### Step 2: Rearranging the equation to find the ratio of \(b\) to \(a\).
From the equation \(b \cos \theta = a \sin \theta\), we can express:
\[ \frac{b}{a} = \frac{\sin \theta}{\cos \theta} = \tan \theta \]

### Step 3: Substitute \(b\) in terms of \(a\) and \(\tan \theta\).
Thus, we have:
\[ b = a \tan \theta \]

### Step 4: Substitute \(b\) into the expression we need to evaluate.
We need to evaluate:
\[ \frac{a}{\sec 2\theta} + \frac{b}{\csc 2\theta} \]
Substituting \(b = a \tan \theta\):
\[ \frac{a}{\sec 2\theta} + \frac{a \tan \theta}{\csc 2\theta} \]

### Step 5: Simplify the expression using trigonometric identities.
Recall that:
\[ \sec 2\theta = \frac{1}{\cos 2\theta} \quad \text{and} \quad \csc 2\theta = \frac{1}{\sin 2\theta} \]
Thus, we can rewrite the expression as:
\[ a \cos 2\theta + a \tan \theta \sin 2\theta \]

### Step 6: Use the identities for \(\tan\) and \(\sin\).
Using the identity \(\tan \theta = \frac{\sin \theta}{\cos \theta}\):
\[ = a \cos 2\theta + a \cdot \frac{\sin \theta}{\cos \theta} \cdot \sin 2\theta \]

### Step 7: Substitute the identities for \(\sin 2\theta\) and \(\cos 2\theta\).
Using the double angle formulas:
\[ \sin 2\theta = 2 \sin \theta \cos \theta \quad \text{and} \quad \cos 2\theta = \cos^2 \theta - \sin^2 \theta \]
We can rewrite:
\[ = a (\cos^2 \theta - \sin^2 \theta) + a \cdot \frac{\sin \theta}{\cos \theta} \cdot (2 \sin \theta \cos \theta) \]
This simplifies to:
\[ = a (\cos^2 \theta - \sin^2 \theta) + 2a \sin^2 \theta \]

### Step 8: Combine like terms.
Combining these terms gives:
\[ = a \cos^2 \theta + a \sin^2 \theta = a (\cos^2 \theta + \sin^2 \theta) \]
Using the Pythagorean identity \(\cos^2 \theta + \sin^2 \theta = 1\):
\[ = a \cdot 1 = a \]

### Final Result:
Thus, the value of the expression is:
\[ \boxed{a} \]