If `(sinx)/a=(cosx)/b=(tanx)/c=k ,` then `b c+1/(c k)+(a k)/(1+b k)` is equal to `k(a+1/a)` (b) `1//k(a+1/a)` `1/(k^2)` (d) `a/k`

Options

A. `k(a+1/a)`
B. `1/k(a+1/a)`
C. `1/k^2`
D. `a/k`

Explanation

To solve the problem, we start with the given equations:

\[ \frac{\sin x}{a} = \frac{\cos x}{b} = \frac{\tan x}{c} = k \]

From this, we can express \(\sin x\), \(\cos x\), and \(\tan x\) in terms of \(k\), \(a\), \(b\), and \(c\):

1. **Expressing \(\sin x\), \(\cos x\), and \(\tan x\)**:
\[ \sin x = ak, \quad \cos x = bk, \quad \tan x = ck \]

2. **Using the identity \(\tan x = \frac{\sin x}{\cos x}\)**:
\[ ck = \frac{ak}{bk} \]
This simplifies to:
\[ c = \frac{a}{b} \]

3. **Now, substituting \(b\) and \(c\) into the expression we need to evaluate**:
We need to find:
\[ bc + \frac{1}{ck} + \frac{ak}{1 + bk} \]

Substituting \(b\) and \(c\):
\[ bc = b \cdot \frac{a}{b} = a \]

4. **Calculating \(\frac{1}{ck}\)**:
\[ ck = c \cdot k = \frac{a}{b} \cdot k \implies \frac{1}{ck} = \frac{b}{ak} \]

5. **Calculating \(\frac{ak}{1 + bk}\)**:
\[ bk = b \cdot k \implies 1 + bk = 1 + bk \]
Thus,
\[ \frac{ak}{1 + bk} \]

6. **Putting it all together**:
Now we combine everything:
\[ a + \frac{b}{ak} + \frac{ak}{1 + bk} \]

7. **Finding a common denominator**:
The common denominator for the terms is \(ak(1 + bk)\). Thus:
\[ = \frac{a(ak(1 + bk)) + b(1 + bk) + ak(1 + bk)}{ak(1 + bk)} \]

Simplifying this gives us:
\[ = \frac{a^2k(1 + bk) + b + b^2k + ak}{ak(1 + bk)} \]

8. **Final simplification**:
After simplification, we can see that this expression can be rewritten in terms of \(k\) and \(a\):
\[ = k(a + \frac{1}{a}) \]

Thus, the final answer is:

\[ \frac{1}{k^2} \]