If cos (theta+phi)=frac{3}{5} and sin (theta- | vedclass

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(A) Given $\cos (	heta+\phi) = \frac{3}{5}$, since $0 < 	heta, \phi < \frac{\pi}{4}$, we have $	an (	heta+\phi) = \frac{4}{3}$.Given $\sin (	heta

Vedclass · Accepted Answer

$\frac{16}{63}$

Vedclass · Answer

(A) Given $\cos (	heta+\phi) = \frac{3}{5}$, since $0 Given $\sin (	heta-\phi) = \frac{5}{13}$, we have $	an (	heta-\phi) = \frac{5}{12}$.We know that $2	heta = (	heta+\phi) + (	heta-\phi)$.Therefore, $	an (2	heta) = 	an ((	heta+\phi) + (	heta-\phi)) = \frac{	an (	heta+\phi) + 	an (	heta-\phi)}{1 - 	an (	heta+\phi) 	an (	heta-\phi)}$.Substituting the values: $	an (2	heta) = \frac{\frac{4}{3} + \frac{5}{12}}{1 - (\frac{4}{3} 	imes \frac{5}{12})} = \frac{\frac{16+5}{12}}{1 - \frac{20}{36}} = \frac{\frac{21}{12}}{\frac{16}{36}} = \frac{21}{12} 	imes \frac{36}{16} = \frac{21 	imes 3}{16} = \frac{63}{16}$.Thus, $\cot (2	heta) = \frac{1}{	an (2	heta)} = \frac{16}{63}$.

If $\cos (\theta+\phi)=\frac{3}{5}$ and $\sin (\theta-\phi)=\frac{5}{13}$, where $0 < \theta, \phi < \frac{\pi}{4}$, then $\cot (2 \theta)$ has the value:

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