If tan theta_1 = k cot theta_2,then frac{cos | vedclass

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(B) Given that $	an 	heta_1 = k \cot 	heta_2$. Since $\cot 	heta_2 = \frac{1}{	an 	heta_2}$,we have $	an 	heta_1 = \frac{k}{	an 	heta_2}$,wh

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$\frac{1-k}{1+k}$

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(B) Given that $	an 	heta_1 = k \cot 	heta_2$. Since $\cot 	heta_2 = \frac{1}{	an 	heta_2}$,we have $	an 	heta_1 = \frac{k}{	an 	heta_2}$,which implies $	an 	heta_1 	an 	heta_2 = k$. Now,consider the expression $\frac{\cos (	heta_1 + 	heta_2)}{\cos (	heta_1 - 	heta_2)}$. Using the expansion formulas,we get $\frac{\cos 	heta_1 \cos 	heta_2 - \sin 	heta_1 \sin 	heta_2}{\cos 	heta_1 \cos 	heta_2 + \sin 	heta_1 \sin 	heta_2}$. Dividing the numerator and denominator by $\cos 	heta_1 \cos 	heta_2$,we obtain $\frac{1 - 	an 	heta_1 	an 	heta_2}{1 + 	an 	heta_1 	an 	heta_2}$. Substituting $	an 	heta_1 	an 	heta_2 = k$,the expression becomes $\frac{1-k}{1+k}$.

If $\tan \theta_1 = k \cot \theta_2$,then $\frac{\cos (\theta_1 + \theta_2)}{\cos (\theta_1 - \theta_2)} = $

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