In Fig. A M=M C and C is a right angle, then ^2 - ^2 is equal to

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MCQ

In Fig. $A M=M C$ and $\angle C$ is a right angle, then $\sin ^2 \alpha-\cos ^2 \alpha$ is equal to

A
$\frac{4 b^2-3 a^2}{5 a^2-4 b^2}$
✓
$\frac{5 a^2-4 b^2}{4 b^2-3 a^2}$
C
$\frac{4 a^2-5 b^2}{3 b^2-4 a^2}$
D
$\frac{3 b^2-4 a^2}{4 a^2-5 b^2}$

✓

Answer

Correct option: B.

$\frac{5 a^2-4 b^2}{4 b^2-3 a^2}$

(B)$\frac{5 a^2-4 b^2}{4 b^2-3 a^2}$
Applying Pythagoras Theorem in right triangle $A B C$, we obtain
$
A B^2=A C^2+B C^2 \Rightarrow b^2=a^2+B C^2 \Rightarrow B C=\sqrt{b^2-a^2}
$
Thus, in right triangle $B C M$, we obtain: $B C=\sqrt{b^2-a^2}$ and $C M=a / 2$.
Applying Pythagoras Theorem in $\triangle B C M$, we obtain
$
B M^2=B C^2+C M^2 \Rightarrow B M^2=b^2-a^2+\frac{a^2}{4}=\frac{4 b^2-3 a^2}{4} \Rightarrow B M=\frac{\sqrt{4 b^2-3 a^2}}{2}
$
In $\triangle B C M$, we obtain
$
\begin{aligned}
& \sin \alpha=\frac{C M}{B M}=\frac{a / 2}{\frac{\sqrt{4 b^2-3 a^2}}{2}}=\frac{a}{\sqrt{4 b^2-3 a^2}} \text { and } \cos \alpha=\frac{B C}{B M}=\frac{2 \sqrt{b^2-a^2}}{\sqrt{4 b^2-3 a^2}} \\
\therefore \quad & \sin ^2 \alpha-\cos ^2 \alpha=\frac{a^2}{4 b^2-3 a^2}-\frac{4\left(b^2-a^2\right)}{4 b^2-3 a^2}=\frac{5 a^2-4 b^2}{4 b^2-3 a^2}
\end{aligned}
$

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