By considering $1^{\prime}=0.0175$, the approximate value of $\cot 45^{\circ} 2^{\prime}$ is

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Aicharya · Accepted Answer

Let $f(x)=\cot x$ Since $1^{\prime}=0.0175 \Rightarrow 2^{\prime}=0.035$ Now, $f^{\prime}(x)=-\operatorname{cosec}^2 x$ By approximation, we get $\begin{aligned} & f(a+h)=f(a)+h f^{\prime}(a) \ & \Rightarrow f\left(45^{\circ}+2^{\prime}ight)=f\left(45^{\circ}ight)+2^{\prime} f^{\prime}\left(45^{\circ}ight) \ & =\cot \left(45^{\circ}ight)-2^{\prime} \operatorname{cosec}^2\left(45^{\circ}ight)\end{aligned}$ $\begin{aligned} & =1-0.035 	imes 2=1-0.07 \ & \Rightarrow \cot \left(45^{\circ} 2^{\prime}ight)=0.93\end{aligned}$

By considering $1^{\prime}=0.0175$, the approximate value of $\cot 45^{\circ} 2^{\prime}$ is

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