MHT CET · Maths · Inverse Trigonometric Functions
If \(x=\operatorname{cosec}\left(\tan ^{-1}\left(\cos \left(\cot ^{-1}\left(\sec \left(\sin ^{-1} a\right)\right)\right)\right)\right)\), \(\mathrm{a} \in[0,1]\)
- A \(x^2-\mathrm{a}^2=3\)
- B \(x^2+\mathrm{a}^2=3\)
- C \(x^2-\mathrm{a}^2=2\)
- D \(x^2+\mathrm{a}^2=2\)
Answer & Solution
Correct Answer
(B) \(x^2+\mathrm{a}^2=3\)
Step-by-step Solution
Detailed explanation
\( x=\operatorname{cosec}(\tan ^{-1}(\cos (\cot ^{-1}(\sec (\sin ^{-1} \mathrm{a}))))) \)
\( =\operatorname{cosec}(\tan ^{-1}(\cos (\cot ^{-1}(\sec (\sec ^{-1} \frac{1}{\sqrt{1-\mathrm{a}^2}}))))) \)
\( =\operatorname{cosec}(\tan ^{-1}(\cos (\cot ^{-1}(\frac{1}{\sqrt{1-\mathrm{a}^2}})))) \)
\( =\operatorname{cosec}(\tan ^{-1}(\cos (\cos ^{-1} \frac{1}{\sqrt{2-\mathrm{a}^2}}))) \)
\( =\operatorname{cosec}(\tan ^{-1}(\frac{1}{\sqrt{2-\mathrm{a}^2}})) \)
\( =\operatorname{cosec}(\operatorname{cosec}^{-1}(\sqrt{3-\mathrm{a}^2})) \)
\( \therefore \quad x=\sqrt{3-\mathrm{a}^2} \)
\( \therefore \quad x^2+\mathrm{a}^2=3\)
\( =\operatorname{cosec}(\tan ^{-1}(\cos (\cot ^{-1}(\sec (\sec ^{-1} \frac{1}{\sqrt{1-\mathrm{a}^2}}))))) \)
\( =\operatorname{cosec}(\tan ^{-1}(\cos (\cot ^{-1}(\frac{1}{\sqrt{1-\mathrm{a}^2}})))) \)
\( =\operatorname{cosec}(\tan ^{-1}(\cos (\cos ^{-1} \frac{1}{\sqrt{2-\mathrm{a}^2}}))) \)
\( =\operatorname{cosec}(\tan ^{-1}(\frac{1}{\sqrt{2-\mathrm{a}^2}})) \)
\( =\operatorname{cosec}(\operatorname{cosec}^{-1}(\sqrt{3-\mathrm{a}^2})) \)
\( \therefore \quad x=\sqrt{3-\mathrm{a}^2} \)
\( \therefore \quad x^2+\mathrm{a}^2=3\)
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