Question

If p =(tan (3^n+1 theta)-tan theta) and q=sigma n r=0 sin (3^r theta)÷cos (3^r+1 theta) then

Answer 1

SOLUTION

GIVEN

$\displaystyle \sf{p = ( \tan ({ 3}^{n + 1} \theta) -\tan \theta \: ) \: \: \: \: and \: \: \: q = \sum\limits_{r=0}^{n} \: \frac{ \sin ({ 3}^{r} \theta)}{ \cos ({ 3}^{r + 1} \theta)} }$

TO DETERMINE

The relation between p and q

EVALUATION

Here it is given that

$\displaystyle \sf{p = ( \tan ({ 3}^{n + 1} \theta) -\tan \theta \: ) }$

$\displaystyle \sf{q = \sum\limits_{r=0}^{n} \: \frac{ \sin ({ 3}^{r} \theta)}{ \cos ({ 3}^{r + 1} \theta)} }$

$\displaystyle \sf{Let \: \: \alpha = { 3}^{r} \theta }$

Now

$\displaystyle \sf{\frac{ \sin ({ 3}^{r} \theta)}{ \cos ({ 3}^{r + 1} \theta)} }$

$\displaystyle \sf{ = \frac{ \sin ({ 3}^{r} \theta)}{ \cos (3.{ 3}^{r } \theta)} }$

$\displaystyle \sf{ = \frac{ \sin \alpha }{ \cos 3 \alpha } }$

$\displaystyle \sf{ = \frac{ 2\sin \alpha \cos \alpha}{ 2\cos 3 \alpha \cos \alpha} }$

$\displaystyle \sf{ = \frac{ \sin 2\alpha }{ 2\cos 3 \alpha \cos \alpha} }$

$\displaystyle \sf{ = \frac{ \sin (3\alpha - \alpha)}{ 2\cos 3 \alpha \cos \alpha} }$

$\displaystyle \sf{ = \frac{ \sin 3\alpha \cos \alpha - \cos 3\alpha \sin \alpha}{ 2\cos 3 \alpha \cos \alpha} }$

$\displaystyle \sf{ = \frac{1}{2} \bigg[\frac{ \sin 3\alpha \cos \alpha - \cos 3\alpha \sin \alpha}{ 2\cos 3 \alpha \cos \alpha} \bigg]}$

$\displaystyle \sf{ = \frac{1}{2} \bigg[\frac{ \sin 3\alpha \cos \alpha }{ \cos 3 \alpha \cos \alpha} - \frac{ \cos 3\alpha \sin \alpha}{ \cos 3 \alpha \cos \alpha}\bigg]}$

$\displaystyle \sf{ = \frac{1}{2} \bigg[ \tan 3\alpha - \tan \alpha\bigg]}$

$\displaystyle \sf{ = \frac{1}{2} \bigg[ \tan { 3}^{r + 1} \theta - \tan { 3}^{r} \theta\bigg]}$

Thus we get

$\displaystyle \sf{q = \sum\limits_{r=0}^{n} \: \frac{ \sin ({ 3}^{r} \theta)}{ \cos ({ 3}^{r + 1} \theta)} }$

$\displaystyle \sf{ \implies \: q = \sum\limits_{r=0}^{n} \:\frac{1}{2} \bigg[ \tan { 3}^{r + 1} \theta - \tan { 3}^{r} \theta\bigg]}$

$\displaystyle \sf{ \implies \: q = \frac{1}{2} \bigg[ \tan { 3}^{n + 1} \theta - \tan { 3}^{0} \theta\bigg]}$

$\displaystyle \sf{ \implies \: q = \frac{1}{2} \bigg[ \tan { 3}^{n + 1} \theta - \tan \theta\bigg]}$

$\displaystyle \sf{ \implies \: q = \frac{1}{2} \times p}$

$\displaystyle \sf{ \implies \: q = \frac{p}{2} }$

$\displaystyle \sf{ \implies \: p = 2q}$

FINAL ANSWER

Hence the required relationship is

$\boxed{ \: \displaystyle \sf{ \pink{ \: p = 2q}} \: \: }$

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