A) \[3\sqrt{3}-6\]
B) \[\sqrt{3}-2\]
C) \[4\sqrt{3}-7\]
D) \[4\sqrt{3}+7\]
Correct Answer: C
Solution :
[c]| Given, |
| \[\sin \alpha +\cos \beta =\frac{1}{\sqrt{2}}\] ? (i) |
| \[\cos \alpha +\sin \beta =\frac{\sqrt{2}}{\sqrt{3}}\] ? (ii) |
| On subtracting Eq. (ii) from Eq. (i), we get \[(\sin \alpha -\sin \beta )+(\cos \beta -\cos \alpha )=\frac{1}{\sqrt{2}}-\frac{\sqrt{2}}{\sqrt{3}}\] |
| \[2\cos \frac{\alpha +\beta }{2}\sin \left( \frac{\alpha -\beta }{2} \right)+2\sin \frac{\alpha +\beta }{2}\] |
| \[\sin \frac{\alpha -\beta }{2}=\frac{\sqrt{3}-2}{\sqrt{6}}\] |
| \[2\sin \left( \frac{\alpha +\beta }{2} \right)\left[ \cos \frac{\alpha +\beta }{2}+\sin \frac{\alpha +\beta }{2} \right]\] | ||
| \[=\frac{\sqrt{3}-2}{\sqrt{6}}\] | ? (iii) | |
| On adding Eqs. (i) and (ii), we get \[2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}+2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}\] \[=\frac{\sqrt{3}+2}{\sqrt{6}}\] | ||
| \[2\cos \left( \frac{\alpha -\beta }{2} \right)\left[ \sin \frac{\alpha +\beta }{2}+\cos \frac{\alpha +\beta }{2} \right]\] | ||
| \[=\frac{\sqrt{3}+2}{\sqrt{6}}\] | ? (iv) | |
| On dividing Eq. (iii) by Eq. (iv), we get \[\tan \left( \frac{\alpha -\beta }{2} \right)=\frac{\sqrt{3}-2}{\sqrt{3}+2}=4\sqrt{3}-7\] | ||
You need to login to perform this action.
You will be redirected in
3 sec
