The value of lim _{theta rightarrow 0} frac{ | vedclass

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Question

(A) Given that,$ \lim _{	heta ightarrow 0} \frac{1-\cos 4 	heta}{1-\cos 6 	heta} $.Using the trigonometric identity $ 1 - \cos 2x = 2 \sin^2 x $,

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$ \frac{4}{9} $

Vedclass · Answer

(A) Given that,$ \lim _{	heta ightarrow 0} \frac{1-\cos 4 	heta}{1-\cos 6 	heta} $.Using the trigonometric identity $ 1 - \cos 2x = 2 \sin^2 x $,we can rewrite the expression as:$ \lim _{	heta ightarrow 0} \frac{2 \sin^2(2 	heta)}{2 \sin^2(3 	heta)} = \lim _{	heta ightarrow 0} \frac{\sin^2(2 	heta)}{\sin^2(3 	heta)} $.Multiplying and dividing by $ (2 	heta)^2 $ and $ (3 	heta)^2 $ respectively:$ \lim _{	heta ightarrow 0} \left( \frac{\sin(2 	heta)}{2 	heta} ight)^2 	imes \left( \frac{3 	heta}{\sin(3 	heta)} ight)^2 	imes \frac{(2 	heta)^2}{(3 	heta)^2} $.Since $ \lim _{x ightarrow 0} \frac{\sin x}{x} = 1 $,the expression simplifies to:$ 1^2 	imes 1^2 	imes \frac{4 	heta^2}{9 	heta^2} = \frac{4}{9} $.

The value of $ \lim _{\theta \rightarrow 0} \frac{1-\cos 4 \theta}{1-\cos 6 \theta} $ is

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