7option (b).Any good method to solve this q?
\hspace{-16}$Let $\mathbf{A=\sin \sqrt{2}-\sin \sqrt{3}}$ and $\mathbf{B=\cos \sqrt{3}-\cos \sqrt{2}}$\\\\ Then Correct options is,::::\\\\ $\mathbf{(a)\;\; A>0\;\;,B>0}$\\\\ $\mathbf{(b)\;\; A>0\;\;,B<0}$\\\\ $\mathbf{(c)\;\; A<0\;\;,B>0}$\\\\ $\mathbf{(d)\;\; A<0\;\;,B<0}$\\\\
options::(b)
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7 Answers
1√2 , √3 = 1.414 , 1.732 lie in pi/2 to pi { 1.57 , 3.14}
so from graph it is clear that sin iand cos are dec. functions
=> so obviously answer is b.
1there is a long method.
firstly √3=1.732 and √2=1.414
Ï€/2=1.57
3Ï€/4=2.355
So Ï€2<√3<3Ï€4
and Ï€4<√2<Ï€2
1.732-1.57=0.162
1.57-1.414=0.156
So,sin1.414 > sin1.732
A>0
and obviously B<0
1sin and cos both are decrease function w.r.t √2,√3.so that why option B is right
9A=2sin(√3-√22)cos(√3+√22)
hence A>0
B=2sin(√2-√32)cos(√3+√22)
hence B<0
