Question

please proof this......​

Answer 1

Let,

$\longrightarrow\sf{P(n):\dfrac{1}{\sin(2x)}+\dfrac{1}{\sin(2^2x)}+\,\dots\,+\dfrac{1}{\sin(2^nx)}=\cot x-\cot(2^nx)\quad\forall n\in\mathbb{N}}$

Let us check whether P(1) is true.

$\longrightarrow\sf{\dfrac{1}{\sin(2x)}=\dfrac{\sin x}{\sin x\sin(2x)}}$

$\longrightarrow\sf{\dfrac{1}{\sin(2x)}=\dfrac{\sin (2x-x)}{\sin x\sin(2x)}}$

$\longrightarrow\sf{\dfrac{1}{\sin(2x)}=\dfrac{\sin(2x)\cos x-\cos(2x)\sin x}{\sin x\sin(2x)}}$

$\longrightarrow\sf{\dfrac{1}{\sin(2x)}=\dfrac{\cos x}{\sin x}-\dfrac{\cos(2x)}{\sin(2x)}}$

$\longrightarrow\sf{\dfrac{1}{\sin(2x)}=\cot x-\cot(2x)}$

Hence P(1) is true. So assume P(k) is true.

$\longrightarrow\sf{P(k):\dfrac{1}{\sin(2x)}+\dfrac{1}{\sin(2^2x)}+\,\dots\,+\dfrac{1}{\sin(2^kx)}=\cot x-\cot(2^kx)\quad\forall k\in\mathbb{N}}$

Consider P(k + 1).

$\longrightarrow\sf{P(k+1):\dfrac{1}{\sin(2x)}+\dfrac{1}{\sin(2^2x)}+\,\dots\,+\dfrac{1}{\sin(2^{k+1}x)}=\cot x-\cot(2^{k+1}x)\quad\!\!\!\forall k\in\mathbb{N}}$

Let's check whether P(k + 1) is true.

From P(k),

$\begin{aligned}\longrightarrow\ \ &\sf{\dfrac{1}{\sin(2x)}+\dfrac{1}{\sin(2^2x)}+\,\dots\,+\dfrac{1}{\sin(2^kx)}+\dfrac{1}{\sin(2^{k+1}x)}}\\\\=\ \ &\sf{\cot x-\cot(2^kx)+\dfrac{1}{\sin(2^{k+1}x)}}\end{aligned}$

$\begin{aligned}\longrightarrow\ \ &\sf{\dfrac{1}{\sin(2x)}+\dfrac{1}{\sin(2^2x)}+\,\dots\,+\dfrac{1}{\sin(2^kx)}+\dfrac{1}{\sin(2^{k+1}x)}}\\\\=\ \ &\sf{\cot x-\dfrac{\cos(2^kx)}{\sin(2^kx)}+\dfrac{1}{\sin(2^{k+1}x)}}\end{aligned}$

$\begin{aligned}\longrightarrow\ \ &\sf{\dfrac{1}{\sin(2x)}+\dfrac{1}{\sin(2^2x)}+\,\dots\,+\dfrac{1}{\sin(2^kx)}+\dfrac{1}{\sin(2^{k+1}x)}}\\\\=\ \ &\sf{\cot x-\dfrac{\cos(2^kx)\sin(2^{k+1}x)-\sin(2^kx)}{\sin(2^kx)\sin(2^{k+1}x)}}\end{aligned}$

$\begin{aligned}\longrightarrow\ \ &\sf{\dfrac{1}{\sin(2x)}+\dfrac{1}{\sin(2^2x)}+\,\dots\,+\dfrac{1}{\sin(2^kx)}+\dfrac{1}{\sin(2^{k+1}x)}}\\\\=\ \ &\sf{\cot x-\dfrac{\cos(2^kx)\sin(2\cdot2^{k}x)-\sin(2^kx)}{\sin(2^kx)\sin(2^{k+1}x)}}\end{aligned}$

Since $\sf{\sin(2A)=2\sin A\cos A,}$

$\begin{aligned}\longrightarrow\ \ &\sf{\dfrac{1}{\sin(2x)}+\dfrac{1}{\sin(2^2x)}+\,\dots\,+\dfrac{1}{\sin(2^kx)}+\dfrac{1}{\sin(2^{k+1}x)}}\\\\=\ \ &\sf{\cot x-\dfrac{2\sin(2^kx)\cos^2(2^kx)-\sin(2^kx)}{\sin(2^kx)\sin(2^{k+1}x)}}\end{aligned}$

$\begin{aligned}\longrightarrow\ \ &\sf{\dfrac{1}{\sin(2x)}+\dfrac{1}{\sin(2^2x)}+\,\dots\,+\dfrac{1}{\sin(2^kx)}+\dfrac{1}{\sin(2^{k+1}x)}}\\\\=\ \ &\sf{\cot x-\dfrac{2\cos^2(2^kx)-1}{\sin(2^{k+1}x)}}\end{aligned}$

Since $\sf{\cos(2A)=2\cos^2A-1,}$

$\begin{aligned}\longrightarrow\ \ &\sf{\dfrac{1}{\sin(2x)}+\dfrac{1}{\sin(2^2x)}+\,\dots\,+\dfrac{1}{\sin(2^kx)}+\dfrac{1}{\sin(2^{k+1}x)}}\\\\=\ \ &\sf{\cot x-\dfrac{\cos(2\cdot2^kx)}{\sin(2^{k+1}x)}}\end{aligned}$

$\begin{aligned}\longrightarrow\ \ &\sf{\dfrac{1}{\sin(2x)}+\dfrac{1}{\sin(2^2x)}+\,\dots\,+\dfrac{1}{\sin(2^kx)}+\dfrac{1}{\sin(2^{k+1}x)}}\\\\=\ \ &\sf{\cot x-\dfrac{\cos(2^{k+1}x)}{\sin(2^{k+1}x)}}\end{aligned}$

$\longrightarrow\sf{\dfrac{1}{\sin(2x)}+\dfrac{1}{\sin(2^2x)}+\,\dots\,+\dfrac{1}{\sin(2^{k+1}x)}=\cot x-\cot(2^{k+1}x)}$

Thus P(k + 1) is true whenever P(k) is true.

∴ P(n) holds true ∀n ∈ N.

Hence Proved!

Answer 2

Its a verified answer please check this