Question

If $x = \frac{sec \frac{x}{2} \cdot sec \frac{x}{4} \cdot sec \frac{x}{8} \cdots}{cosec x}$, then $\frac{1}{2^2}sec^2\frac{x}{2} + \frac{1}{2^4}sec^2\frac{x}{4} + \frac{1}{2^6}sec^2\frac{x}{8} + \cdots =$

• A. $\frac{x^2(1+\frac{1-sin^2x}{sin^2x})-1}{x^2}$
• B. $(\frac{1+tan^2x}{1-cos^2x})\cdot x^2$
• C. $tan^2x - \frac{1}{2+13sec^2x}$
• D. $(\frac{1+sin^2x}{1+sec^2x})\frac{x}{1+x}$

Answer 1

Answer: (A)

$\frac{x^2(1+\frac{1-sin^2x}{sin^2x})-1}{x^2}$

Answer 2

The given equation is $x = \frac{\sec \frac{x}{2} \cdot \sec \frac{x}{4} \cdot \sec \frac{x}{8} \cdots}{\cosec x}$.

We can write this as $x = \left( \prod_{n=1}^{\infty} \sec \frac{x}{2^n} \right) \sin x$. Using the identity $\sec \theta = \frac{1}{\cos \theta}$, we have $\prod_{n=1}^{\infty} \sec \frac{x}{2^n} = \frac{1}{\prod_{n=1}^{\infty} \cos \frac{x}{2^n}}$. We know the infinite product formula $\prod_{n=1}^{\infty} \cos \frac{x}{2^n} = \frac{\sin x}{x}$ for $x \neq 0$. Substituting this into the expression for $x$: $x = \left( \frac{1}{\frac{\sin x}{x}} \right) \sin x = \frac{x}{\sin x} \cdot \sin x = x$. This confirms that the given equation is an identity for $x \neq 0$.

We need to find the value of the series $S = \frac{1}{2^2}\sec^2\frac{x}{2} + \frac{1}{2^4}\sec^2\frac{x}{4} + \frac{1}{2^6}\sec^2\frac{x}{8} + \cdots = \sum_{n=1}^{\infty} \frac{1}{4^n} \sec^2 \frac{x}{2^n}$.

Consider the identity $\prod_{n=1}^{\infty} \cos \frac{x}{2^n} = \frac{\sin x}{x}$. Taking the natural logarithm of both sides, we get: $\ln \left( \prod_{n=1}^{\infty} \cos \frac{x}{2^n} \right) = \ln \left( \frac{\sin x}{x} \right)$ $\sum_{n=1}^{\infty} \ln \left( \cos \frac{x}{2^n} \right) = \ln (\sin x) - \ln x$.

Differentiate both sides with respect to $x$: $\frac{d}{dx} \left( \sum_{n=1}^{\infty} \ln \left( \cos \frac{x}{2^n} \right) \right) = \frac{d}{dx} (\ln (\sin x) - \ln x)$ $\sum_{n=1}^{\infty} \frac{d}{dx} \left( \ln \left( \cos \frac{x}{2^n} \right) \right) = \frac{\cos x}{\sin x} - \frac{1}{x}$ $\sum_{n=1}^{\infty} \frac{1}{\cos \frac{x}{2^n}} \cdot \left( -\sin \frac{x}{2^n} \right) \cdot \frac{1}{2^n} = \cot x - \frac{1}{x}$ $\sum_{n=1}^{\infty} -\frac{1}{2^n} \tan \frac{x}{2^n} = \cot x - \frac{1}{x}$ $\sum_{n=1}^{\infty} \frac{1}{2^n} \tan \frac{x}{2^n} = \frac{1}{x} - \cot x$.

Differentiate both sides again with respect to $x$: $\frac{d}{dx} \left( \sum_{n=1}^{\infty} \frac{1}{2^n} \tan \frac{x}{2^n} \right) = \frac{d}{dx} \left( \frac{1}{x} - \cot x \right)$ $\sum_{n=1}^{\infty} \frac{d}{dx} \left( \frac{1}{2^n} \tan \frac{x}{2^n} \right) = -\frac{1}{x^2} - (-\cosec^2 x)$ $\sum_{n=1}^{\infty} \frac{1}{2^n} \cdot \sec^2 \frac{x}{2^n} \cdot \frac{1}{2^n} = \cosec^2 x - \frac{1}{x^2}$ $\sum_{n=1}^{\infty} \frac{1}{4^n} \sec^2 \frac{x}{2^n} = \cosec^2 x - \frac{1}{x^2}$.

The series we need to evaluate is $S = \sum_{n=1}^{\infty} \frac{1}{4^n} \sec^2 \frac{x}{2^n}$. So, $S = \cosec^2 x - \frac{1}{x^2}$.

Now let's check the options. Option A: $\frac{x^2(1+\frac{1-\sin^2x}{\sin^2x})-1}{x^2} = \frac{x^2(1+\frac{\cos^2x}{\sin^2x})-1}{x^2} = \frac{x^2(1+\cot^2x)-1}{x^2} = \frac{x^2\cosec^2x - 1}{x^2} = \cosec^2x - \frac{1}{x^2}$. This matches our result.

The final answer is $\cosec^2 x - \frac{1}{x^2}$. Let's check if this is equivalent to Option A. Option A is $\frac{x^2(1+\frac{1-\sin^2x}{\sin^2x})-1}{x^2}$. We have $1 - \sin^2 x = \cos^2 x$. So, $1 + \frac{1-\sin^2 x}{\sin^2 x} = 1 + \frac{\cos^2 x}{\sin^2 x} = 1 + \cot^2 x = \cosec^2 x$. Substituting this into Option A: $\frac{x^2(\cosec^2 x)-1}{x^2} = \frac{x^2 \cosec^2 x}{x^2} - \frac{1}{x^2} = \cosec^2 x - \frac{1}{x^2}$. So, Option A is indeed equal to the value of the series.