Question

$\frac{1}{\tan 3\mathrm{A}-\tan \mathrm{A}}-\frac{1}{\cot 3\mathrm{A}-\cot \mathrm{A}}=K\cot 2\mathrm{A}$. Find the value of $K$.

Answer 1

Find the value of $K$.

It is given that

$$\begin{gathered} \frac{1}{\tan 3\mathrm{A}-\tan \mathrm{A}}-\frac{1}{\cot 3\mathrm{A}-\cot \mathrm{A}}=K\cot 2\mathrm{A} \\ \Rightarrow \frac{1}{{\displaystyle \frac{\sin 3\mathrm{A}}{\cos 3\mathrm{A}}}-{\displaystyle \frac{\sin \mathrm{A}}{\cos \mathrm{A}}}}-\frac{1}{{\displaystyle \frac{\cos 3\mathrm{A}}{\sin 3\mathrm{A}}-\frac{\cos \mathrm{A}}{\sin \mathrm{A}}}}=K\cot 2\mathrm{A}\left[\begin{array}{c}\because \tan \mathrm{A}=\frac{\sin \mathrm{A}}{\cos \mathrm{A}},\\ \cot \mathrm{A}=\frac{\cos \mathrm{A}}{\sin \mathrm{A}}\end{array}\right] \\ \Rightarrow \frac{1}{{\displaystyle \frac{\sin 3\mathrm{A}\cos \mathrm{A}-\sin \mathrm{A}\cos 3\mathrm{A}}{\cos 3\mathrm{A}\cos \mathrm{A}}}}-\frac{1}{{\displaystyle \frac{\cos 3\mathrm{A}\sin \mathrm{A}-\cos \mathrm{A}\sin 3\mathrm{A}}{\sin 3\mathrm{A}\sin \mathrm{A}}}}=K\cot 2\mathrm{A} \\ \Rightarrow \frac{\cos 3\mathrm{A}\cos \mathrm{A}}{{\displaystyle \sin 3\mathrm{A}\cos \mathrm{A}-\sin \mathrm{A}\cos 3\mathrm{A}}}-\frac{\sin 3\mathrm{A}\sin \mathrm{A}}{{\displaystyle \cos 3\mathrm{A}\sin \mathrm{A}-\cos \mathrm{A}\sin 3\mathrm{A}}}=K\cot 2\mathrm{A} \\ \Rightarrow \frac{\cos 3\mathrm{A}\cos \mathrm{A}}{{\displaystyle \sin \left(3\mathrm{A}-\mathrm{A}\right)}}-\frac{\sin 3\mathrm{A}\sin \mathrm{A}}{{\displaystyle \sin \left(\mathrm{A}-3\mathrm{A}\right)}}=K\cot 2\mathrm{A}\left[\because \sin \left(\mathrm{A}+\mathrm{B}\right)=\sin \mathrm{A}\cos \mathrm{B}-\sin \mathrm{B}\cos \mathrm{A}\right] \\ \Rightarrow \frac{\cos 3\mathrm{A}\cos \mathrm{A}}{{\displaystyle \sin 2\mathrm{A}}}-\frac{\sin 3\mathrm{A}\sin \mathrm{A}}{{\displaystyle \sin \left(-2\mathrm{A}\right)}}=\mathrm{K}\cot 2\mathrm{A} \\ \Rightarrow \frac{\cos 3\mathrm{A}\cos \mathrm{A}}{{\displaystyle \sin 2\mathrm{A}}}+\frac{\sin 3\mathrm{A}\sin \mathrm{A}}{{\displaystyle \sin 2\mathrm{A}}}=K\cot 2\mathrm{A}\left[\because \sin \left(-\mathrm{A}\right)=-\sin \left(\mathrm{A}\right)\right] \\ \Rightarrow \frac{\cos 3\mathrm{A}\cos \mathrm{A}+\sin 3\mathrm{A}\sin \mathrm{A}}{{\displaystyle \sin 2\mathrm{A}}}=K\cot 2\mathrm{A} \\ \Rightarrow \frac{\cos \left(3\mathrm{A}-\mathrm{A}\right)}{{\displaystyle \sin 2\mathrm{A}}}=K\cot 2\mathrm{A}\left[\because \cos \left(\mathrm{A}-\mathrm{B}\right)=\cos \mathrm{A}\cos \mathrm{B}-\sin \mathrm{A}\sin \mathrm{B}\right] \\ \Rightarrow \frac{\cos 2\mathrm{A}}{{\displaystyle \sin 2\mathrm{A}}}=K\cot 2\mathrm{A} \\ \Rightarrow \cot 2\mathrm{A}=K\cot 2\mathrm{A} \\ \Rightarrow K=\frac{\cot 2\mathrm{A}}{\cot 2\mathrm{A}} \\ \therefore K=1 \end{gathered}$$

Hence, the value of $K$ is $1$.