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If $\tan ^{2} \mathrm{~B}=\frac{1-\sin \mathrm{A}}{1+\sin \mathrm{A}}$ then what is the value of $\mathrm{A}+2 \mathrm{~B}$ ?
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The correct answer is:
$\frac{\pi}{2}$
Given equation is $\tan ^{2} \mathrm{~B}=\frac{1-\sin \mathrm{A}}{1+\sin \mathrm{A}}$
$\Rightarrow$ Applying componendo and dividendo $\frac{1+\tan ^{2} \mathrm{~B}}{1-\tan ^{2} \mathrm{~B}}=\frac{2}{2 \sin \mathrm{A}}$
$\Rightarrow \sin \mathrm{A}=\frac{1-\tan ^{2} \mathrm{~B}}{1+\tan ^{2} \mathrm{~B}} \Rightarrow \sin \mathrm{A}=\cos 2 \mathrm{~B}$
$\Rightarrow \sin A=\sin \left(\frac{\pi}{2}-2 \mathrm{~B}\right)$
$\Rightarrow \mathrm{A}=\frac{\pi}{2}-2 \mathrm{~B} \Rightarrow \mathrm{A}+2 \mathrm{~B}=\frac{\pi}{2}$
$\Rightarrow$ Applying componendo and dividendo $\frac{1+\tan ^{2} \mathrm{~B}}{1-\tan ^{2} \mathrm{~B}}=\frac{2}{2 \sin \mathrm{A}}$
$\Rightarrow \sin \mathrm{A}=\frac{1-\tan ^{2} \mathrm{~B}}{1+\tan ^{2} \mathrm{~B}} \Rightarrow \sin \mathrm{A}=\cos 2 \mathrm{~B}$
$\Rightarrow \sin A=\sin \left(\frac{\pi}{2}-2 \mathrm{~B}\right)$
$\Rightarrow \mathrm{A}=\frac{\pi}{2}-2 \mathrm{~B} \Rightarrow \mathrm{A}+2 \mathrm{~B}=\frac{\pi}{2}$
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