Area Question 47
Question 47
- Find the area of the region bounded by the curve $C: y=\tan x$, tangent drawn to $C$ at $x=\pi / 4$ and the $X$-axis.
$(1988,5 \mathrm{M})$
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Solution:
- Given, $y=\tan x \Rightarrow \frac{d y}{d x}=\sec ^{2} x$
$$ \therefore \quad \frac{d y}{d x}{ }_{x=\frac{\pi}{4}}=2 $$
Hence, equation of tangent at $A \frac{\pi}{4}, 1$ is
$$ \frac{y-1}{x-\pi / 4}=2 \Rightarrow y-1=2 x-\frac{\pi}{2} $$
$$ \Rightarrow \quad(2 x-y)=\frac{\pi}{2}-1 $$
$\therefore$ Required area is $O A B O$
$$ \begin{aligned} & =\int_{0}^{\pi / 4}(\tan x) d x-\text { area of } \triangle A L B \ & =[\log |\sec x|]_{0}^{\pi / 4}-\frac{1}{2} \cdot B L \cdot A L \ & =\log \sqrt{2}-\frac{1}{2} \frac{\pi}{4}-\frac{\pi-2}{4} \cdot 1 \ & =\log \sqrt{2}-\frac{1}{4} \text { sq unit } \end{aligned} $$
