If cos x -2- cos x -22- cos x -23- - - cos x -2n-sin x -2nsinx-2 - Then1-2tan x -2-1-22tan x -22-1-2 n tan x -22 iA- 1-2ntanx-2n-tan xB- x-2n- xC- x-x-2nD- 1-2nx-2n- x

The correct option is D Since, nbsp; 1/2tan(x/2) = nbsp; 1/2cot(x/2) cotx ∴ nbsp; 1/2^2tan(x/2^2) = nbsp; 1/2^2cot(x/2^2) nbsp; 1/2cot( x/2) 1/2 ...

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Standard XII

Mathematics

Properties Derived from Trigonometric Identities

If cos x /2, ...

Question

If cos $\frac{x}{2}$ , cos $\frac{x}{2^{2}}$ , cos $\frac{x}{2^{3}}$ .............cos $\frac{x}{2^{n}}$ = $\frac{s i n x}{2^{n} s i n \frac{x}{2^{n}}}$ Then

$\frac{1}{2}$ tan $\frac{x}{2}$ + $\frac{1}{2^{2}}$ tan $\frac{x}{2^{2}}$ + ............. $\frac{1}{2^{n}}$ tan $\frac{x}{2^{n}}$ is

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Solution

The correct option is B

Since, $\frac{1}{2}$ tan $(\frac{x}{2})$ = $\frac{1}{2}$ cot $(\frac{x}{2})$ - cotx

∴ $\frac{1}{2^{2}}$ tan $(\frac{x}{2^{2}})$ = $\frac{1}{2^{2}}$ cot $(\frac{x}{2^{2}})$ - $\frac{1}{2}$ cot( $\frac{x}{2}$ )

$\frac{1}{2^{3}}$ tan $(\frac{x}{2^{3}})$ = $\frac{1}{2^{3}}$ cot $(\frac{x}{2^{3}})$ - $\frac{1}{2^{2}}$ cot $(\frac{x}{2^{2}})$

$\frac{1}{2^{n}}$ tan $(\frac{x}{2^{n}})$ = $\frac{1}{2^{n}}$ cot $(\frac{x}{2^{n}})$ - $\frac{1}{2^{n - 1}}$ cot $(\frac{x}{2^{n - 1}})$

Adding we get

$\frac{1}{2}$ tan( $\frac{x}{2}$ ) + $\frac{1}{2^{2}}$ tan $(\frac{x}{2^{2}})$ +..................+ $\frac{1}{2^{n}}$ tan $(\frac{x}{2^{n}})$

= $\frac{1}{2^{n}}$ cot $(\frac{x}{2^{n}})$ -cot x

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Similar questions

lim n \to \infty (\frac{1}{2} tan \frac{x}{2} + \frac{1}{2^{2}} tan \frac{x}{2^{2}} + \frac{1}{2^{3}} tan \frac{x}{2^{3}} + . . . . . . + \frac{1}{2^{n}} tan \frac{x}{2^{n}})

is equal to.

Prove that $\frac{1}{2} t a n (\frac{x}{2}) + \frac{1}{4} t a n (\frac{x}{4}) + . . . + \frac{1}{2^{n}} t a n (\frac{x}{2^{n}}) = \frac{1}{2^{n}} c o t + (\frac{x}{2^{n}}) - c o t x$ for all $n ϵ N$ and $0 < x < \frac{x}{2}$ .

cos \frac{x}{2} \times cos \frac{x}{2^{2}} \times cos \frac{x}{2^{3}} . . . . . . . . . . \times cos \frac{x}{2^{n}}

A \to \infty

\frac{sin 2^{n} A}{2^{n} sin A}

Q. Given that

n \prod n = 1 c o s \frac{x}{2^{n}} = \frac{sin x}{2^{n} sin (\frac{x}{2^{n}})}

and

f (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} {lim}_{n \to \infty} \sum_{n = 1}^{n} \frac{1}{2^{n}} tan (\frac{x}{2^{n}}), & x \in (0, π) - {\frac{π}{2}} \frac{2}{π} & x = \frac{π}{2} \end{matrix}

Then which one of the following is true?

L e t f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} (1 + | c o s x |)^{\frac{a b}{| c o s x |}}, & n π < x < (2 n + 1) \frac{π}{2} e^{a} . e^{b}, & x = (2 n + 1) \frac{π}{2} e^{\frac{c o t 2 x}{c o t 8 x}}, & (2 n + 1) \frac{π}{2} < x < (n + 1) π \end{matrix} I f f (x) i s c o n t i n u o u s i n ((n π), (n + 1) π, t h e n)

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If cos x2, cos x22, cos x23.............cos x2n = sinx2nsinx2nThen 12tan x2 + 122tan x22 + ............. 12ntan x2n is

The correct option is B Since, 12tan(x2) = 12cot(x2) - cotx ∴ 122tan(x22) = 122cot(x22) - 12cot( x2) 123 tan(x23) = 123cot(x23) - 122cot(x22) 12n tan(x2n) = 12ncot(x2n) - 12n−1cot(x2n−1) Adding we get 12 tan( x2) + 122tan(x22) +..................+ 12ntan(x2n) = 12n cot(x2n) -cot x

If cos $\frac{x}{2}$ , cos $\frac{x}{2^{2}}$ , cos $\frac{x}{2^{3}}$ .............cos $\frac{x}{2^{n}}$ = $\frac{s i n x}{2^{n} s i n \frac{x}{2^{n}}}$ Then

$\frac{1}{2}$ tan $\frac{x}{2}$ + $\frac{1}{2^{2}}$ tan $\frac{x}{2^{2}}$ + ............. $\frac{1}{2^{n}}$ tan $\frac{x}{2^{n}}$ is