If cosx-2 cosx-22 cosx-22 - to -sin x-x then 1-22 2x-2 1-22 2x-22- -

The correct option is B log cosx/2+log cosx/2^2+......log sin x log x ⇒ 1/2 tanx/2 1/2^2tanx/2^2...=cot x 1/x ⇒1/2^2 sec^2x/2+1/2^4sec^2x/2^2+...=cox ec^2x 1/ ...

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Standard XII

Mathematics

Implicit Differentiation

If cosx/2 cos...

Question

$If cos (\frac{x}{2}) c o s (\frac{x}{2^{2}}) c o s (\frac{x}{2^{3}}) . . . . . . . . . . . . . . . t o \infty = \frac{s i n x}{x} t h e n \frac{1}{2^{2}} s e c^{2} (\frac{x}{2}) \frac{1}{2^{4}} s e c^{2} (\frac{x}{2^{2}}) + . . . . . . . =$

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$cos~ec^{2}x+\frac{1}{x^2}$

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Solution

The correct option is B

$l o g c o s \frac{x}{2} + l o g c o s \frac{x}{2^{2}} + . . . . . . l o g s i n x - l o g x \Rightarrow \frac{- 1}{2} t a n \frac{x}{2} - \frac{1}{2^{2}} t a n \frac{x}{2^{2}} . . . = c o t x - \frac{1}{x} \Rightarrow \frac{1}{2^{2}} s e c^{2} \frac{x}{2} + \frac{1}{2^{4}} s e c^{2} \frac{x}{2^{2}} + . . . = c o x e c^{2} x - \frac{1}{x^{2}}$

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Similar questions

Q. Given that

cos \frac{x}{2} cos \frac{x}{4} cos \frac{x}{8} \dots = \frac{sin x}{x}

, then expression

\frac{1}{2^{2}} {sec}^{2} \frac{x}{2} + \frac{1}{2^{4}} {sec}^{2} \frac{x}{4} + \dots

equals

$If cos (\frac{x}{2}) c o s (\frac{x}{2^{2}}) c o s (\frac{x}{2^{3}}) . . . . . . . . . . . . . . . t o \infty = \frac{s i n x}{x} t h e n \frac{1}{2^{2}} s e c^{2} (\frac{x}{2}) \frac{1}{2^{4}} s e c^{2} (\frac{x}{2^{2}}) + . . . . . . . =$

Q. Given that

cos \frac{x}{2} . cos \frac{x}{4} . cos \frac{x}{8} . . . . = \frac{sin x}{x}

, prove

\frac{1}{2^{2}} {sec}^{2} \frac{x}{2} + \frac{1}{2^{4}} {sec}^{2} \frac{x}{4} + . . . . = {csc}^{2} x - \frac{1}{x^{2}}

lim x \to \frac{π}{2} (lim x \to \infty cos \frac{x}{2} cos \frac{x}{2^{2}} cos \frac{x}{2^{3}} . . . . . . cos \frac{x}{2^{n}})

equals to

Q. If

sin x + cos x = \sqrt{2} cos x

, then cosx - sin x is equal to

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If cos(x2)cos(x22)cos(x23)...............to ∞=sin xxthen 122sec2(x2)124sec2(x22)+.......=

The correct option is B log cosx2+log cosx22+......log sin x−log x⇒−12 tanx2−122tanx22...=cot x−1x⇒122 sec2x2+124sec2x22+...=cox ec2x−1x2

$If cos (\frac{x}{2}) c o s (\frac{x}{2^{2}}) c o s (\frac{x}{2^{3}}) . . . . . . . . . . . . . . . t o \infty = \frac{s i n x}{x} t h e n \frac{1}{2^{2}} s e c^{2} (\frac{x}{2}) \frac{1}{2^{4}} s e c^{2} (\frac{x}{2^{2}}) + . . . . . . . =$