Question

If $\displaystyle\sf T_n = sin^n\:\theta+cos^n\:\theta$ , prove that $\displaystyle\sf \dfrac{T_3 - T_5}{T_1} = \dfrac{T_5 - T_7}{T_3}$​

Answer 1

For solution refer to attachments

Answer 2

$\large\underline{\sf{Given- }}$

$\rm :\longmapsto\:T_n = {sin}^{n}\theta \: + {cos}^{n}\theta$

$\large\underline{\sf{To\:prove-}}$

$\rm :\longmapsto\:\dfrac{T_3 - T_5}{T_1} = \dfrac{T_5 - T_7}{T_3}$

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\boxed{ \bf \: 1 - {sin}^{2}x = {cos}^{2}x}$

$\boxed{ \bf \: 1 - {cos}^{2}x = {sin}^{2}x}$

$\large\underline{\bf{Solution-}}$

Given that,

$\rm :\longmapsto\:T_n = {sin}^{n}\theta \: + {cos}^{n}\theta$

Thus,

$\rm :\longmapsto\:T_1 = {sin}^{}\theta \: + {cos}^{}\theta$

$\rm :\longmapsto\:T_3 = {sin}^{3}\theta \: + {cos}^{3}\theta$

$\rm :\longmapsto\:T_5 = {sin}^{5}\theta \: + {cos}^{5}\theta$

$\rm :\longmapsto\:T_7 = {sin}^{7}\theta \: + {cos}^{7}\theta$

Now,

Consider,

$\rm :\longmapsto\:\dfrac{T_3 - T_5}{T_1}$

$\sf \:  =  \:  \: \dfrac{({sin}^{3}\theta + {cos}^{3}\theta) - ( {sin}^{5}\theta + {cos}^{5} \theta)}{sin\theta + cos\theta}$

$\sf \:  =  \:  \: \dfrac{{sin}^{3}\theta + {cos}^{3}\theta - {sin}^{5}\theta - {cos}^{5} \theta}{sin\theta + cos\theta}$

$\sf \:  =  \:  \: \dfrac{{sin}^{3}\theta - {sin}^{5}\theta + {cos}^{3}\theta - {cos}^{5} \theta}{sin\theta + cos\theta}$

$\sf \:  =  \:  \: \dfrac{{sin}^{3}\theta(1 - {sin}^{2}\theta) + {cos}^{3}\theta(1 - {cos}^{2})\theta}{sin\theta + cos\theta}$

$\sf \:  =  \:  \: \dfrac{{sin}^{3}\theta {cos}^{2}\theta+{cos}^{3}\theta{sin}^{2}\theta}{sin\theta + cos\theta}$

$\sf \:  =  \:  \: \dfrac{{sin}^{2}\theta {cos}^{2}\theta({cos}^{}\theta + {sin}^{}\theta)}{sin\theta + cos\theta}$

$\sf \:  =  \:  \: {sin}^{2}\theta + {cos}^{2}\theta$

$\bf\implies \:\:\dfrac{T_3 - T_5}{T_1} = {sin}^{2}\theta {cos}^{2}\theta - - (1)$

Consider,

$\rm :\longmapsto\:\dfrac{T_5 - T_7}{T_3}$

$\sf \:  =  \:  \: \dfrac{({sin}^{5}\theta + {cos}^{5}\theta) - ( {sin}^{7}\theta + {cos}^{7} \theta)}{ {sin}^{3}\theta + {cos}^{3}\theta}$

$\sf \:  =  \:  \: \dfrac{{sin}^{5}\theta + {cos}^{5}\theta-{sin}^{7}\theta - {cos}^{7} \theta}{ {sin}^{3}\theta + {cos}^{3}\theta}$

$\sf \:  =  \:  \: \dfrac{{sin}^{5}\theta - {sin}^{7}\theta + {cos}^{5}\theta - {cos}^{7} \theta}{ {sin}^{3}\theta + {cos}^{3}\theta}$

$\sf \:  =  \:  \: \dfrac{{sin}^{5}\theta(1 - {sin}^{2}\theta) + {cos}^{5}\theta(1 - {cos}^{2}\theta)}{ {sin}^{3}\theta + {cos}^{3}\theta}$

$\sf \:  =  \:  \: \dfrac{{sin}^{5}\theta{cos}^{2}\theta+ {cos}^{5}\theta{sin}^{2}\theta}{ {sin}^{3}\theta + {cos}^{3}\theta}$

$\sf \:  =  \:  \: \dfrac{{sin}^{2}\theta{cos}^{2}\theta \: ({cos}^{3}\theta + {sin}^{3}\theta)}{ {sin}^{3}\theta + {cos}^{3}\theta}$

$\sf \:  =  \:  \: {sin}^{2}\theta {cos}^{2}\theta$

$\bf\implies \:\:\dfrac{T_5 - T_7}{T_3} = {sin}^{2}\theta {cos}^{2}\theta - - (2)$

Hence,

From equation (1) and equation (2), concluded that

$\bf :\longmapsto\:\dfrac{T_3 - T_5}{T_1} = \dfrac{T_5 - T_7}{T_3}$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

$\boxed{ \bf \: 1 + {tan}^{2}x = {sec}^{2}x}$

$\boxed{ \bf \: {tan}^{2}x = {sec}^{2}x - 1}$

$\boxed{ \bf \: {cot}^{2}x = {cosec}^{2}x - 1}$

$\boxed{ \bf \: 1 + {cot}^{2}x = {cosec}^{2}x}$

$\boxed{ \bf \:{cosec}^{2}x - {cot}^{2}x = 1 }$

$\boxed{ \bf \:{sec}^{2}x - {tan}^{2}x = 1 }$

$\boxed{ \bf \:{sin}^{2}x + {cos}^{2}x = 1 }$