MHT CET · Maths · Inverse Trigonometric Functions
If \(\tan ^{-1}(x+2)+\tan ^{-1}(x-2)-\tan ^{-1}\left(\frac{1}{2}\right)=0\), then one value of \(x\) is
- A -1
- B \(\frac{1}{2}\)
- C 1
- D 2
Answer & Solution
Correct Answer
(C) 1
Step-by-step Solution
Detailed explanation
\(\tan ^{-1}(x+2)+\tan ^{-1}(x-2)-\tan ^{-1}\left(\frac{1}{2}\right)=0 \)
\( \Rightarrow \tan ^{-1}\left(\frac{(x+2)+(x-2)}{1-(x+2)(x-2)}\right)-\tan ^{-1}\left(\frac{1}{2}\right)=0 \)
\( \Rightarrow \tan ^{-1}\left(\frac{2 x}{5-x^2}\right)-\tan ^{-1}\left(\frac{1}{2}\right)=0 \)
\( \Rightarrow \tan ^{-1}\left(\frac{\frac{2 x}{5-x^2}-\frac{1}{2}}{1+\frac{2 x}{5-x^2} \times \frac{1}{2}}\right)=0 \)
\( \Rightarrow \tan ^{-1}\left(\frac{x^2+4 x-5}{-x^2+x+5}\right)=0 \)
\( \therefore \quad x^2+4 x-5=0 \)
\( \therefore \quad x=-5 \text { or } x=1\)
\( \Rightarrow \tan ^{-1}\left(\frac{(x+2)+(x-2)}{1-(x+2)(x-2)}\right)-\tan ^{-1}\left(\frac{1}{2}\right)=0 \)
\( \Rightarrow \tan ^{-1}\left(\frac{2 x}{5-x^2}\right)-\tan ^{-1}\left(\frac{1}{2}\right)=0 \)
\( \Rightarrow \tan ^{-1}\left(\frac{\frac{2 x}{5-x^2}-\frac{1}{2}}{1+\frac{2 x}{5-x^2} \times \frac{1}{2}}\right)=0 \)
\( \Rightarrow \tan ^{-1}\left(\frac{x^2+4 x-5}{-x^2+x+5}\right)=0 \)
\( \therefore \quad x^2+4 x-5=0 \)
\( \therefore \quad x=-5 \text { or } x=1\)
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