Prove the following trigonometric identities.
Question:

Prove the following trigonometric identities.

If $\mathrm{T}_{n}=\sin ^{n} \theta+\cos ^{n} \theta$, prove that $\frac{\mathrm{T}_{3}-\mathrm{T}_{5}}{\mathrm{~T}_{1}}=\frac{\mathrm{T}_{5}-\mathrm{T}_{7}}{\mathrm{~T}_{3}}$

Solution:

In the given question, we are given $T_{n}=\sin ^{n} \theta+\cos ^{n} \theta$

We need to prove $\frac{T_{3}-T_{5}}{T_{1}}=\frac{T_{5}-T_{7}}{T_{3}}$

Here L.H.S is

$\frac{T_{3}-T_{5}}{T_{1}}=\frac{\left(\sin ^{3} \theta+\cos ^{3} \theta\right)-\left(\sin ^{5} \theta+\cos ^{5} \theta\right)}{(\sin \theta+\cos \theta)}$

Now, solving the L.H.S, we get

$\frac{\left(\sin ^{3} \theta+\cos ^{3} \theta\right)-\left(\sin ^{5} \theta+\cos ^{5} \theta\right)}{(\sin \theta+\cos \theta)}=\frac{\sin ^{3} \theta-\sin ^{5} \theta+\cos ^{3} \theta-\cos ^{5} \theta}{\sin \theta+\cos \theta}$

$=\frac{\sin ^{3} \theta\left(1-\sin ^{2} \theta\right)+\cos ^{3} \theta\left(1-\cos ^{2} \theta\right)}{\sin \theta+\cos \theta}$

Further using the property $\sin ^{2} \theta+\cos ^{2} \theta=1$, we get

$\cos ^{2} \theta=1-\sin ^{2} \theta$

$\sin ^{2} \theta=1-\cos ^{2} \theta$

So,

$\frac{\sin ^{3} \theta\left(1-\sin ^{2} \theta\right)+\cos ^{3} \theta\left(1-\cos ^{2} \theta\right)}{\sin \theta+\cos \theta}=\frac{\sin ^{3} \theta \cos ^{2} \theta+\cos ^{3} \theta \sin ^{2} \theta}{\sin \theta+\cos \theta}$

$=\frac{\sin ^{2} \theta \cos ^{2} \theta(\sin \theta+\cos \theta)}{\sin \theta+\cos \theta}$

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

Now, solving the R.H.S, we get

$\frac{T_{5}-T_{7}}{T_{3}}=\frac{\left(\sin ^{5} \theta+\cos ^{5} \theta\right)-\left(\sin ^{7} \theta+\cos ^{7} \theta\right)}{\left(\sin ^{3} \theta+\cos ^{3} \theta\right)}$

So,

$\frac{\left(\sin ^{5} \theta+\cos ^{5} \theta\right)-\left(\sin ^{7} \theta+\cos ^{7} \theta\right)}{\left(\sin ^{3} \theta+\cos ^{3} \theta\right)}=\frac{\sin ^{5} \theta-\sin ^{7} \theta+\cos ^{5} \theta-\cos ^{7} \theta}{\sin ^{3} \theta+\cos ^{3} \theta}$

$=\frac{\sin ^{5} \theta\left(1-\sin ^{2} \theta\right)+\cos ^{5} \theta\left(1-\cos ^{2} \theta\right)}{\sin ^{3} \theta+\cos ^{3} \theta}$

Further using the property $\sin ^{2} \theta+\cos ^{2} \theta=1$, we get,

$\cos ^{2} \theta=1-\sin ^{2} \theta$

$\sin ^{2} \theta=1-\cos ^{2} \theta$

So,

$\frac{\sin ^{5} \theta\left(1-\sin ^{2} \theta\right)+\cos ^{5} \theta\left(1-\cos ^{2} \theta\right)}{\sin ^{3} \theta+\cos ^{3} \theta}=\frac{\sin ^{5} \theta \cos ^{2} \theta+\cos ^{5} \theta \sin ^{2} \theta}{\sin ^{3} \theta+\cos ^{3} \theta}$

$=\frac{\sin ^{2} \theta \cos ^{2} \theta\left(\sin ^{3} \theta+\cos ^{3} \theta\right)}{\sin ^{3} \theta+\cos ^{3} \theta}$

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

Hence proved.