Question

If \(A = \left[ \begin{array}{cc}1 & -tan\frac{θ}{2} \\ tan\frac{θ}{2} & 1\end{array} \right]\) and \(B = \left[ \begin{array}{cc}1 & tan\frac{θ}{2} \\ -tan\frac{θ}{2} & 1\end{array} \right]\) then \(A^{-1}B\) is equal to

• A. \([\begin{array}{cc}cosθ & sinθ \\ -sinθ & cosθ\end{array}]\)
• B. \([\begin{array}{cc}cosθ & -sinθ \\ sinθ & cosθ\end{array}]\)
• C. \([\begin{array}{cc}sinθ & -cosθ \\ cosθ & sinθ\end{array}]\)
• D. \([\begin{array}{cc}cosθ & sinθ \\ sinθ & cosθ\end{array}]\)

Hint:

Compare the leading coefficients of numerator and denominator, since degrees match.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The numerator has degree \(19 + 11 = 30\), the same as the denominator \((6x - 5)^{30}\). The limit at infinity is the ratio of leading coefficients.

Step 2: Leading coefficients:
\[ \lim_{x\to\infty}\frac{(2x)^{19}(3x)^{11}}{(6x)^{30}} = \frac{2^{19}\cdot 3^{11}}{6^{30}} = \frac{2^{19}\cdot 3^{11}}{2^{30}\cdot 3^{30}} = 2^{-11}\cdot 3^{-19} \]

Step 3: Compare with \(2^a 3^b\):
So \(a = -11\) and \(b = -19\), and \(a + b = -30\).

Final Answer:
\(a + b = -30\), option (A). \[ \boxed{-30} \]