If `asectheta+btantheta+c=0andpsectheta+qtantheta+r=0`, then `(br-qc)^(2)-(pc-ar)^(2)=?`

Options

A. a) `(bp-aq)`
B. b) `(aq-bp)^(2)`
C. c) `(aq+bp)^(2)`
D. d) none of these

Answer

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

1. \( a \sec \theta + b \tan \theta + c = 0 \)
2. \( p \sec \theta + q \tan \theta + r = 0 \)

From these equations, we can express \( c \) and \( r \) in terms of \( \sec \theta \) and \( \tan \theta \):

### Step 1: Express \( c \) in terms of \( \sec \theta \) and \( \tan \theta \)
From the first equation:
\[ c = - (a \sec \theta + b \tan \theta) \]

### Step 2: Express \( r \) in terms of \( \sec \theta \) and \( \tan \theta \)
From the second equation:
\[ r = - (p \sec \theta + q \tan \theta) \]

### Step 3: Substitute \( c \) and \( r \) into the expression \( (br - qc)^2 - (pc - ar)^2 \)
We need to evaluate:
\[ (br - qc)^2 - (pc - ar)^2 \]

Substituting the values of \( c \) and \( r \):
\[ br = b(- (p \sec \theta + q \tan \theta)) = -bp \sec \theta - bq \tan \theta \]
\[ qc = q(- (a \sec \theta + b \tan \theta)) = -qa \sec \theta - qb \tan \theta \]

Now, substituting these into \( br - qc \):
\[ br - qc = (-bp \sec \theta - bq \tan \theta) - (-qa \sec \theta - qb \tan \theta) \]
\[ = (-bp + qa) \sec \theta + (-bq + qb) \tan \theta \]
\[ = (qa - bp) \sec \theta + 0 \tan \theta \]
Thus, we have:
\[ (br - qc)^2 = (qa - bp)^2 \sec^2 \theta \]

### Step 4: Evaluate \( pc - ar \)
Now, we calculate:
\[ pc = p(- (a \sec \theta + b \tan \theta)) = -pa \sec \theta - pb \tan \theta \]
\[ ar = a(- (p \sec \theta + q \tan \theta)) = -ap \sec \theta - aq \tan \theta \]

Now substituting these into \( pc - ar \):
\[ pc - ar = (-pa \sec \theta - pb \tan \theta) - (-ap \sec \theta - aq \tan \theta) \]
\[ = (-pa + ap) \sec \theta + (-pb + aq) \tan \theta \]
\[ = 0 \sec \theta + (aq - pb) \tan \theta \]
Thus, we have:
\[ (pc - ar)^2 = (aq - pb)^2 \tan^2 \theta \]

### Step 5: Combine the two results
Now we substitute back into our expression:
\[ (br - qc)^2 - (pc - ar)^2 = (qa - bp)^2 \sec^2 \theta - (aq - pb)^2 \tan^2 \theta \]

### Step 6: Factor out the common term
Factoring out \( (qa - bp)^2 \) and \( (aq - pb)^2 \):
\[ = (qa - bp)^2 \sec^2 \theta - (aq - pb)^2 \tan^2 \theta \]

Using the identity \( \sec^2 \theta - \tan^2 \theta = 1 \):
\[ = (qa - bp)^2 \cdot 1 \]

### Final Result
Thus, we conclude:
\[ (br - qc)^2 - (pc - ar)^2 = (qa - bp)^2 \]