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

Question

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

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Isabelle 8 months 2021-10-03T09:30:55+00:00 1 Answer 0 views 0

Answers ( )

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    2021-10-03T09:32:39+00:00

    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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