If frac{cos (theta_1+theta_2)}{cos (theta_1-t | vedclass

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(A) Given,$\frac{\cos (	heta_1+	heta_2)}{\cos (	heta_1-	heta_2)}+\frac{\cos (	heta_3-	heta_4)}{\cos (	heta_3+	heta_4)}=0$. Dividing the numera

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$1$

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(A) Given,$\frac{\cos (	heta_1+	heta_2)}{\cos (	heta_1-	heta_2)}+\frac{\cos (	heta_3-	heta_4)}{\cos (	heta_3+	heta_4)}=0$. Dividing the numerator and denominator of the first term by $\cos 	heta_1 \cos 	heta_2$ and the second term by $\cos 	heta_3 \cos 	heta_4$,we get: $\frac{1-	an 	heta_1 	an 	heta_2}{1+	an 	heta_1 	an 	heta_2} + \frac{1-	an 	heta_3 	an 	heta_4}{1+	an 	heta_3 	an 	heta_4} = 0$. Let $x = 	an 	heta_1 	an 	heta_2$ and $y = 	an 	heta_3 	an 	heta_4$. Then $\frac{1-x}{1+x} + \frac{1-y}{1+y} = 0$. $(1-x)(1+y) + (1-y)(1+x) = 0$. $1 + y - x - xy + 1 + x - y - xy = 0$. $2 - 2xy = 0 \Rightarrow xy = 1$. Thus,$(	an 	heta_1 	an 	heta_2)(	an 	heta_3 	an 	heta_4) = 1$. Therefore,$\cot 	heta_1 \cot 	heta_2 \cot 	heta_3 \cot 	heta_4 = \frac{1}{	an 	heta_1 	an 	heta_2 	an 	heta_3 	an 	heta_4} = \frac{1}{1} = 1$.

If $\frac{\cos (\theta_1+\theta_2)}{\cos (\theta_1-\theta_2)}+\frac{\cos (\theta_3-\theta_4)}{\cos (\theta_3+\theta_4)}=0$,then $\cot \theta_1 \cdot \cot \theta_2 \cdot \cot \theta_3 \cdot \cot \theta_4=$

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