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Prove the following identities:If $ T_{n}=\sin ^{n} \theta+\cos ^{n} \theta $, prove that $ \frac{T_{3}-T_{5}}{T_{1}}=\frac{T_{5}-T_{7}}{T_{3}} $
Given:
\( T_{n}=\sin ^{n} \theta+\cos ^{n} \theta \)
To do:
We have to prove that \( \frac{T_{3}-T_{5}}{T_{1}}=\frac{T_{5}-T_{7}}{T_{3}} \).
Solution:
We know that,
$\sin^2 A+\cos^2 A=1$
$\operatorname{cosec}^2 A-\cot^2 A=1$
$\sec^2 A-\tan^2 A=1$
$\cot A=\frac{\cos A}{\sin A}$
$\tan A=\frac{\sin A}{\cos A}$
$\operatorname{cosec} A=\frac{1}{\sin A}$
$\sec A=\frac{1}{\cos A}$
Therefore,
Let us consider LHS,
$\frac{\mathrm{T}_{3}-\mathrm{T}_{5}}{\mathrm{~T}_{1}}=\frac{\left(\sin ^{3} \theta+\cos ^{3} \theta\right)-\left(\sin ^{5} \theta+\cos ^{5} \theta\right)}{\sin \theta+\cos \theta}$
$=\frac{\sin ^{3} \theta\left(1-\sin ^{2} \theta\right)+\cos ^{3} \theta\left(1-\cos ^{2} \theta\right)}{\sin \theta+\cos \theta}$
$=\frac{\sin ^{3} \theta\cos ^{2} \theta+\cos ^{3} \theta\sin ^{2} \theta}{\sin \theta+\cos \theta}$
$=\frac{\sin ^{2} \theta \cos ^{2} \theta(\sin \theta+\cos \theta)}{\sin \theta+\cos \theta}$
$=\sin ^{2} \theta \cos ^{2} \theta$
Let us consider RHS,
$\frac{T_{5}-T_{7}}{T_{3}}=\frac{\left(\sin ^{5} \theta+\cos ^{5} \theta\right)-\left(\sin ^{7} \theta+\cos ^{7} \theta\right)}{\sin ^{3} \theta+\cos ^{3} \theta}$
$=\frac{\sin ^{5} \theta+\cos ^{5} \theta-\sin ^{7} \theta-\cos ^{7} \theta}{\sin ^{3} \theta+\cos ^{3} \theta}$
$=\frac{\sin ^{5} \theta-\sin ^{7} \theta+\cos ^{5} \theta-\cos ^{7} \theta}{\sin ^{3} \theta+\cos ^{3} \theta}$
$=\frac{\sin ^{5} \theta\left(1-\sin ^{2} \theta\right)+\cos ^{5} \theta\left(1-\cos ^{2} \theta\right)}{\sin ^{3} \theta+\cos ^{3} \theta}$
$=\frac{\sin ^{5} \theta\cos ^{2} \theta+\cos ^{5} \theta\sin ^{2} \theta}{\sin^3 \theta+\cos^3 \theta}$
$=\frac{\sin ^{2} \theta \cos ^{2} \theta(\sin^3 \theta+\cos^3 \theta)}{\sin^3 \theta+\cos^3 \theta}$
$=\sin ^{2} \theta \cos ^{2} \theta$
Here,
LHS $=$ RHS
Hence proved.
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