JEE Mains · Maths · STD 12 - 2. inverse trigonometric function
If \(\tan \mathrm{A}=\frac{1}{\sqrt{x\left(x^2+x+1\right)}}, \tan B=\frac{\sqrt{x}}{\sqrt{x^2+x+1}}\) and \(\tan C=\left(x^{-3}+x^{-2}+x^{-1}\right)^{\frac{1}{2}}, 0 < A, B, C < \frac{\pi}{2}\) then \(A+B\) is equal to :
- A \(\mathrm{C}\)
- B \(\pi-C\)
- C \(2 \pi-C\)
- D \(\frac{\pi}{2}-C\)
Answer & Solution
Correct Answer
(A) \(\mathrm{C}\)
Step-by-step Solution
Detailed explanation
Finding \(\tan (A+B)\) we get \(\Rightarrow \tan (\mathrm{A}+\mathrm{B})=\) \(\frac{\tan A+\tan B}{1-\tan A \tan B}\) \(=\frac{\frac{1}{\sqrt{x\left(x^2+x+1\right)}}+\frac{\sqrt{x}}{\sqrt{x^2+x+1}}}{1-\frac{1}{x^2+x+1}}\)…
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