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Trigonometry – II — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTrigonometry – II4 Marks
Question
In any triangle $\mathrm{ABC}_r \sin \mathrm{A}-\cos \mathrm{B}=\cos \mathrm{C}$. Show that $\angle \mathrm{B}=\frac{\pi}{2}$.
✓

Answer

$ \sin A-\cos B=\cos C$
$ \therefore \sin A=\cos B+\cos C$
$ \therefore 2 \sin \frac{\mathrm{~A}}{2} \cos \frac{\mathrm{~A}}{2}=2 \cos \left(\frac{\mathrm{~B}+\mathrm{C}}{2}\right) \cos \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)$
$ \therefore 2 \sin \frac{\mathrm{~A}}{2} \cos \frac{\mathrm{~A}}{2}=2 \cos \left(\frac{\pi}{2}-\frac{\mathrm{A}}{2}\right) \cos \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)$
$\left[\begin{array}{l}\because \mathrm{A}+\mathrm{B}+\mathrm{C}=\pi, \\ \therefore \frac{\mathrm{B}+\mathrm{C}}{2}=\frac{\pi}{2}-\frac{\mathrm{A}}{2}\end{array}\right]$
$\therefore 2 \sin \frac{\mathrm{~A}}{2} \cos \frac{\mathrm{~A}}{2}=2 \sin \frac{\mathrm{~A}}{2} \cos \left(\frac{\mathrm{~B}-\mathrm{C}}{2}\right)$
$\therefore \cos \frac{\mathrm{A}}{2}=\cos \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right) \ldots\left[\because \sin \frac{\mathrm{A}}{2} \neq 0\right]$
$ \therefore \frac{\mathrm{A}}{2}=\frac{\mathrm{B}-\mathrm{C}}{2}$
$ A=B-C$
$ \text { In } \triangle \mathrm{ABC}$
$ A+B+C=\pi$
$ \therefore B-C+B+C=\pi$
$ \therefore 2 B=\pi$
$ \therefore B=\frac{\pi}{2}$

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