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Inverse Trigonometric Functions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsInverse Trigonometric Functions2 Marks
Question
Evaluate the following:
$\cot\Big(\cot^{-1}\frac{3}{5}\Big)$
✓

Answer

Let $\cot\Big(\cot^{-1}\frac{3}{5}\Big)=\text{y}$ where $\text{y}\in \Big[0,\frac{\pi}{2}\Big]$
$\Rightarrow \cos \text{y}=\frac{3}{5}$
$\cot \Big(\cos^{-1}\frac{3}{5}\Big)=\cot\text{y}$
To fing:
$\Rightarrow \text{As}\ 1+\tan^2\theta=\sec^2\theta$
$\Rightarrow \tan \text{y}=\sqrt{\sec^2\text{y}-1}$ where $\text{y}\in \Big[0, \frac{\pi}{2}\Big]$
$\Rightarrow \frac{1}{\cot\text{y}}=\sqrt{\Big(\frac{1}{\cos^2\text{y}}\Big)-1}$
$\Rightarrow \frac{1}{\cot\text{y}}=\sqrt{\Big(\frac{5}{3}\Big)^2-1}$
$\Rightarrow \frac{1}{\cot\text{y}}=\sqrt{\frac{16}{9}}$
$\Rightarrow \cot\text{y}=\frac{3}{4}$
$\Rightarrow \cot\Big(\cos^{-1}\frac{3}{5}\Big)=\frac{3}{4}$

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