Question

1+cos2A=3cisAsinA...cotA=?

Answer 1

Answer:

Step-by-step explanation:

We have,

$1+\cos(2A)=3\,\cos(A)\,\sin(A)$

$\implies2\cos^2(A)=3\,\cos(A)\,\sin(A)$

$\implies2\cos^2(A)-3\,\cos(A)\,\sin(A)=0$

$\implies\cos(A)\left\{2\,\cos(A)-3\,\sin(A)\right\}=0$

$\implies\cos(A)=0\,\,\,\,\,or\,\,\,\,\,2\,\cos(A)-3\,\sin(A)=0$

$\implies\cos(A)=0\,\,\,\,\,or\,\,\,\,\,2\,\cos(A)=3\,\sin(A)$

$\implies\cos(A)=0\,\,\,\,\,or\,\,\,\,\,\dfrac{\cos(A)}{\sin(A)}=\dfrac{3}{2}$

$\implies\cos(A)=0\,\,\,\,\,or\,\,\,\,\,\cot(A)=\dfrac{3}{2}$

Since  $\cos(A)=0$, so,  $\cot(A)=\dfrac{\cos(A)}{\sin(A)}=0$

So,

$\mapsto\tt{\blue{cot(A)=0\,\,\,\,\,or\,\,\,\,\,cot(A)=\dfrac{3}{2}}}$