Question

If cosectheta=p+q/p-q where p>q>0 then cot(pi/4+theta/2)=​

Answer 1

SOLUTION

GIVEN

$\displaystyle \sf{ \cosec \theta = \frac{p + q}{p - q} }$

TO DETERMINE

$\displaystyle \sf{ \cot \bigg( \frac{\pi}{4} + \frac{ \theta}{2} \bigg) }$

EVALUATION

$\displaystyle \sf{ \cosec \theta = \frac{p + q}{p - q} }$

$\implies \displaystyle \sf{ \sin \theta = \frac{p - q}{p + q} }$

$\implies \displaystyle \sf{ \frac{ 2 \tan \frac{ \theta}{2} }{1 + { \tan}^{2} \frac{ \theta}{2} } = \frac{p + q}{p - q} }$

By Componendo Dividendo Rule

$\implies \displaystyle \sf{ \frac{1 + { \tan}^{2} \frac{ \theta}{2} + 2 \tan \frac{ \theta}{2} }{1 + { \tan}^{2} \frac{ \theta}{2} - 2 \tan \frac{ \theta}{2} } = \frac{2p}{2 q} }$

$\implies \displaystyle \sf{ \frac{ \bigg({1 + { \tan} \frac{ \theta}{2} \bigg)}^{2} }{ \bigg({1 - { \tan} \frac{ \theta}{2} \bigg)}^{2}} = \frac{p}{ q} }$

$\implies \displaystyle \sf{ \frac{ \bigg({1 + { \tan} \frac{ \theta}{2} \bigg)} }{ \bigg({1 - { \tan} \frac{ \theta}{2} \bigg)}} = \pm \: \sqrt{ \frac{p}{ q} }}$

$\implies \displaystyle \sf{ \frac{ \bigg({ \tan \frac{\pi}{4} + { \tan} \frac{ \theta}{2} \bigg)} }{ \bigg({1 - \tan \frac{\pi}{4} { \tan} \frac{ \theta}{2} \bigg)}} = \pm \: \sqrt{ \frac{p}{ q} }}$

$\implies \displaystyle \sf{ { \tan \bigg({ \frac{\pi}{4} + \frac{ \theta}{2} \bigg)} } = \pm \: \sqrt{ \frac{p}{ q} }}$

$\implies \displaystyle \sf{ { \cot \bigg({ \frac{\pi}{4} + \frac{ \theta}{2} \bigg)} } = \pm \: \sqrt{ \frac{q}{p} }}$

FINAL ANSWER

$\boxed{ \: \: \displaystyle \sf{ { \cot \bigg({ \frac{\pi}{4} + \frac{ \theta}{2} \bigg)} } = \pm \: \sqrt{ \frac{q}{p} }} \: \: }$

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