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: Divide by x to the 30:
Divide every bracket by x: the numerator is $(2 - \frac1x)^{19}(3 + \frac2x)^{11}$ and the denominator is $(6 - \frac5x)^{30}$.

Step 2: Let x go to infinity:
The fractions vanish, leaving $\frac{2^{19}3^{11}}{6^{30}} = 2^{-11}3^{-19}$. Hence $a + b = -11 - 19 = -30$ (A).

Final Answer:
-30. \[ \boxed{-30} \]