Q107 / 160 · TS EAMCET 2024 (10 May Shift 1) · Mathematics
The approximate value of $\sec 59^{\circ}$ obtained by taking $1^{\circ}=$ 0.0174 and $\sqrt{3}=1.732$ is
- A1.9849
- B1.8493
- Correct answerC1.9397
- D1.9948
Solution
Let $f(x)=\sec x \Rightarrow f^{\prime}(x)=\sec x \tan x$
Take $a=60^{\circ}$ and $h=-1^{\circ}=-0.0174$ Then $f(a)=\sec 60^{\circ}=2 \Rightarrow f^{\prime}(a)=2 \sqrt{3}$ $\begin{aligned} & \Rightarrow f(a+h) \approx f(a)+h \cdot f^{\prime}(a) \\ & \Rightarrow \quad f\left(59^{\circ}\right) \\ & =\sec 59^{\circ}=\sec 60^{\circ}-(0.0174)(2 \sqrt{3}) \\ & \\ & =2-0.0602736 \Rightarrow \sec 59^{\circ}=1.9397 \end{aligned}$
Take $a=60^{\circ}$ and $h=-1^{\circ}=-0.0174$ Then $f(a)=\sec 60^{\circ}=2 \Rightarrow f^{\prime}(a)=2 \sqrt{3}$ $\begin{aligned} & \Rightarrow f(a+h) \approx f(a)+h \cdot f^{\prime}(a) \\ & \Rightarrow \quad f\left(59^{\circ}\right) \\ & =\sec 59^{\circ}=\sec 60^{\circ}-(0.0174)(2 \sqrt{3}) \\ & \\ & =2-0.0602736 \Rightarrow \sec 59^{\circ}=1.9397 \end{aligned}$
Hence, the correct answer is option: 'C'.