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Question
If A=tan−1(x√32k−x) and B=tan−1(2x−kk√3). Then, A−B is equal to
If A=tan−1(x√32k−x) and B=tan−1(2x−kk√3). Then, A−B is equal to
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Solution
The correct option is C π6
A=tan−1[x√32k−x]B=tan−1[2x−kk√3]
According to inverse trignometric formulae
tan−1A−tan−1B=tan−1[A−B1+AB]
Now, A−B=tan−1[√3x2k−x]−tan−1[2x−k√3k]
=tan−1⎡⎢
⎢
⎢
⎢
⎢⎣√3x2k−x−2x−k√3k1+√3x2k−x×2x−k√3k⎤⎥
⎥
⎥
⎥
⎥⎦
=tan−1[3kx−(2x−k)(2k−x)(2k−x)√3k+√3x(2x−k)]
=tan−1⎡⎢⎣3kx−4kx+2x2+2k2−kx√3(2k2−kx+2x2−kx)⎤⎥⎦
=tan−1⎡⎢⎣−2kx+2x2+2k2√3(2k2+2x2−2kx)⎤⎥⎦
=tan−1[1√3]=π6
A=tan−1[x√32k−x]B=tan−1[2x−kk√3]
According to inverse trignometric formulae
tan−1A−tan−1B=tan−1[A−B1+AB]
Now, A−B=tan−1[√3x2k−x]−tan−1[2x−k√3k]
=tan−1⎡⎢
⎢
⎢
⎢
⎢⎣√3x2k−x−2x−k√3k1+√3x2k−x×2x−k√3k⎤⎥
⎥
⎥
⎥
⎥⎦
=tan−1[3kx−(2x−k)(2k−x)(2k−x)√3k+√3x(2x−k)]
=tan−1⎡⎢⎣3kx−4kx+2x2+2k2−kx√3(2k2−kx+2x2−kx)⎤⎥⎦
=tan−1⎡⎢⎣−2kx+2x2+2k2√3(2k2+2x2−2kx)⎤⎥⎦
=tan−1[1√3]=π6
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