91Ó°ÊÓ

(d) \(\because A+B+C=180^{\circ}\) \(\Rightarrow \quad A=180^{\circ}-(B+C)\) \(\therefore \quad \tan A=\tan \left(180^{\circ}-(B+C)\right)\) \(\quad=-\tan (B+C)=-\left\\{\frac{\tan B+\tan C}{1-\tan B \tan C}\right\\}\) \(\quad=\left\\{\frac{\tan B+\tan C}{\tan B \tan C-1}\right\\}\) Now, \(\because\) A is obtuse \(\therefore \quad \tan A<0\), then \(\tan B+\tan C>0\) \(\therefore \quad \tan B \tan C-1<0\) \(\Rightarrow \quad \tan B \tan C<1\)

(b) \(\cos ^{4} \theta-\sin ^{4} \theta=\left(\cos ^{2} \theta-\sin ^{2} \theta\right)\left(\cos ^{2} \theta+\right.\) \(\left.\sin ^{2} \theta\right)\) \(=\cos 2 \theta=2 \cos ^{2} \theta-1\)

A line through the point \(A(2,0)\), which makes an angle of \(30^{\circ}\) with the positive direction of \(x\)-axis is rotated about \(A\) in clockwise direction through an angle \(15^{\circ}\). The equation of the straight line in the new position is (a) \((2-\sqrt{3}) x-y-4+2 \sqrt{3}=0\) (b) \((2-\sqrt{3}) x+y-4+2 \sqrt{3}=0\) (c) \((2-\sqrt{3}) x-y+4+2 \sqrt{3}=0\) (d) None of these

A rectangle has two opposite vertices at the points \((1,2)\) and \((5,5)\). If the other vertices lie on the line \(x=3\), then the coordinates of the other vertices are (a) \((3,-1),(3,-6)\) (b) \((3,1),(3,5)\) (c) \((3,2),(3,6)\) (d) \((3,1),(3,6)\)