If cot ,theta + tan theta = m and sec theta - | vedclass

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Question

(a) As given 

$\frac{1}{{	an 	heta }} + 	an 	heta = m\, $

$\Rightarrow \,1 + {	an ^2}	heta = m\,	an 	heta $

$ \Rightarrow \,\,{\

Vedclass · Accepted Answer

$m{(m{n^2})^{1/3}} - n{(n{m^2})^{1/3}} = 1$

Vedclass · Answer

(a) As given 

$\frac{1}{{	an 	heta }} + 	an 	heta = m\, $

$\Rightarrow \,1 + {	an ^2}	heta = m\,	an 	heta $

$ \Rightarrow \,\,{\sec ^2}	heta = m\,	an 	heta $&hellip;..$(i) $

and $\sec 	heta - \cos 	heta = n\,\, \Rightarrow \,\,{\sec ^2}	heta - 1 = n\,\sec 	heta $ 

$ \Rightarrow \,\,{	an ^2}	heta = n\,\,\sec 	heta $ 

$ \Rightarrow \,\,{	an ^4}	heta = {n^2}\,{\sec ^2}	heta = {n^2}.\,m\,\,	an 	heta $      {by $(i)$} 

$ \Rightarrow \,\,{	an ^3}	heta  = {n^2}m\,,\,\,\,(\,\,\,	an 	heta  
e 0)$ 

$\Rightarrow \,\,	an 	heta = {({n^2}m)^{1/3}}$&hellip;..$(ii)$ 

Also, ${\sec ^2}	heta = m\,\,	an 	heta = m\,{({n^2}m)^{1/3}}$ {by $(i)$ and $(ii)$} 

$	herefore$ Using the identity ${\sec ^2}	heta - {	an ^2}	heta = 1$ 

$ \Rightarrow \,\,m\,{(m{n^2})^{1/3}} - {({n^2}m)^{2/3}} = 1$ 

$ \Rightarrow \,\,m\,{(m{n^2})^{1/3}} - n\,{(n{m^2})^{1/3}} = 1.$

If $\cot \,\theta + \tan \theta = m$ and $\sec \theta - \cos \theta = n,$ then which of the following is correct

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