AP EAMCET · Maths · Inverse Trigonometric Functions
If \(\cot \left(\operatorname{Cos}^{-1} \mathrm{x}\right)=\sec \left\{\operatorname{Tan}^{-1}\left(\frac{\mathrm{a}}{\sqrt{\mathrm{b}^2-\mathrm{a}^2}}\right)\right\}, \mathrm{b}>\mathrm{a}\), then \(\mathrm{x}=\)
- A \(\frac{\mathrm{b}}{\sqrt{2 \mathrm{~b}^2-\mathrm{a}^2}}\)
- B \(\frac{a}{\sqrt{2 b^2-a^2}}\)
- C \(\frac{\sqrt{b^2-a^2}}{a}\)
- D \(\frac{\sqrt{b^2-a^2}}{b}\)
Answer & Solution
Correct Answer
(A) \(\frac{\mathrm{b}}{\sqrt{2 \mathrm{~b}^2-\mathrm{a}^2}}\)
Step-by-step Solution
Detailed explanation
\( \cot \left(\operatorname{Cos}^{-1} \mathrm{x}\right) = \frac{\mathrm{x}}{\sqrt{1-\mathrm{x}^2}} \) \( \sec \left\{\operatorname{Tan}^{-1}\left(\frac{\mathrm{a}}{\sqrt{\mathrm{b}^2-\mathrm{a}^2}}\right)\right\} = \frac{\mathrm{b}}{\sqrt{\mathrm{b}^2-\mathrm{a}^2}} \)…
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