JEE Mains · Maths · STD 12 - 2. inverse trigonometric function
If \({\cos ^{ - 1}}\left( {\frac{2}{{3x}}} \right) + {\cos ^{ - 1}}\,\left( {\frac{3}{{4x}}} \right) = \frac{\pi }{2},\,x > \frac{3}{4}\) then \(x\) is equal to
- A \(\frac{{\sqrt {145} }}{{12}}\)
- B \(\frac{{\sqrt {145} }}{{10}}\)
- C \(\frac{{\sqrt {146} }}{{12}}\)
- D \(\frac{{\sqrt {145} }}{{11}}\)
Answer & Solution
Correct Answer
(A) \(\frac{{\sqrt {145} }}{{12}}\)
Step-by-step Solution
Detailed explanation
\({\cos ^{ - 1}}\left( {\frac{2}{{3x}}} \right) + {\cos ^{ - 1}}\left( {\frac{3}{{4x}}} \right) = \frac{\pi }{2}\left( {x > \frac{3}{4}} \right)\)…
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