Question

Arrange the following values in the ascending order of their magnitudes

A : lf $A+B+C=\pi$ then

$\frac{\cos A}{\sin B\sin C}+\frac{\cos B}{\sin A\sin C}+\frac{\cos C}{\sin A\sin B}=$
B : If $A+B+C=\pi$ then $\sum \cot A.\cot B=$
C : If $A+B+C=\pi$ then
$\tan 3A+\tan 3B+\tan 3C=2K\tan 3A\tan 3B\tan 3C,thenK=$
D : lf $A+B+C=\pi$ then
$\sec A(cosB\cos C-\sin B\sin C)=$

• A. $D,C,B,A$
• B. $A,B,C,D$
• C. $A,B,D,C$
• D. $D,A,B,C$

Answer 1

Answer: (A)

$D,C,B,A$

$A:$ Given $A+B+C=\pi ......\left(I\right)$

$\frac{\cos A}{\sin B.\sin C}+\frac{\cos B}{\sin A.\sin C}+\frac{\cos C}{\sin A.\sin B}$

$=\frac{\sin A.\cos A+\cos B.\sin B+\cos C.\sin C}{\sin A.\sin B.\sin C}$

$=\frac{1}{2}\frac{2\sin A.\cos B+2.\sin B.\cos B+2.\sin C.\cos C}{\sin A.\sin B.\sin C}$

$=\frac{1}{2}\frac{\sin 2A+\sin 2B+2.\sin C.\cos C}{\sin A.\sin B.\sin C}[\because \sin 2\theta =2\sin \theta .\cos \theta ]$

$=\frac{1}{2}\frac{2\sin \left(\frac{2(A+B)}{2}\right).\cos \left(\frac{2(A-B)}{2}\right)+2.\sin C.\cos C}{\sin A.\sin B.\sin C}$

$[\because \sin C+\sin D=2\sin \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)]$

$=\frac{1}{2}\frac{2.\sin (A+B).\cos (A-B)+2.\sin (\pi -(A+B\left)\right).\cos (\pi -(A+B\left)\right)}{\sin A.\sin B.\sin C}$

$=\frac{1}{2}\frac{2.\sin (A+B).\cos (A-B)+2.\sin (A+B).\cos (A-B)}{\sin A\sin B.\sin C}$

$=\frac{\frac{2.\sin (A+B)}{2}\left[\cos \right(A-B)-\cos (A+B\left)\right]}{\sin A.\sin B.\sin C}$

$=\frac{\sin (A+B)[2.\sin A.\sin B]}{\sin A.\sin B.\sin C}$

$=\frac{2.\sin A.\sin B.\sin C}{\sin A.\sin B.\sin C}$

$=2$

$\therefore A=2$

$B:$ Given $A+B+C=\pi$

$A+B=\pi -C$

$\cot (A+B)=\cot (\pi -C)$

$\frac{\cot A\cot B-1}{\cot A+\cot B}=-\cot C$

$\cot A\cot B-1=-\cot C\cot A-\cot C\cot B$

$\cot A\cot B+\cot B\cot C+\cot C+\cot A=1$

$\therefore B=1$

$C:A+B+C=\pi ,\tan 3A+\tan 3B+\tan 3C=2k\tan 3A\tan 3B\tan 3C$

$3A+3B+3C=3\pi$

$3A+3B=3\pi -3C$

$\tan (3A+3B)=\tan (3\pi -3c)$

$\Rightarrow \frac{\tan 3A+\tan 3B}{1-\tan 3A.\tan 3B}=-\tan 3C$

$\Rightarrow \tan 3A+\tan 3B=-\tan 3C+\tan 3A\tan 3B\tan 3C$

$\Rightarrow \tan 3A+\tan 3B+\tan 3C=\tan 3A\tan 3B\tan 3C$

$=2k\tan 3A.\tan 3B.\tan 3C=\tan 3A\tan 3B\tan C$

$\Rightarrow 2k=1$

$k=\frac{1}{2}$

$D:A+B+C=\pi$

$\sec A(\cos B\cos C-\sin B.\sin C)=\sec A\left(\cos \right(B+C\left)\right)$

$=\sec A\left(\cos \right(\pi -A\left)\right)$

$=\sec A(-\cos A)$

$=-1$

$\therefore D=-1$

From the value

$A=2,B=1,C=k=\frac{1}{2},D=-1$

the ascending order

$-1<\frac{1}{2}<1<2$

$\Rightarrow D<C<B<A$