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MHTCET 16 May 2024 Shift 2 Paper

Question : 136 of 150

Marks: +1, -0

\tan ^{-1}\left(\frac{x+1}{x-1}\right)+\tan ^{-1}\left(\frac{x-1}{x}\right)=\tan ^{-1}(-7)

x

is equal to

Solution:

Concept:

Use the standard formula for the sum of two inverse tangents:

\tan^{-1}A + \tan^{-1}B = \tan^{-1}\left(\frac{A+B}{1-AB}\right)

Explanation:

Given:

\tan^{-1}\left(\frac{x+1}{x-1}\right)+\tan^{-1}\left(\frac{x-1}{x}\right)=\tan^{-1}(-7)

Let

A=\frac{x+1}{x-1}

and

B=\frac{x-1}{x}

Apply the formula:

\tan^{-1}\left[\frac{\frac{x+1}{x-1}+\frac{x-1}{x}}{1-\left(\frac{x+1}{x-1}\right)\left(\frac{x-1}{x}\right)}\right]=\tan^{-1}(-7)

Simplify the numerator:

\frac{x(x+1)+(x-1)^2}{x(x-1)}=\frac{2x^2-x+1}{x(x-1)}

Simplify the denominator:

1-\frac{x+1}{x}=-\frac{1}{x}

So the argument becomes:

\frac{2x^2-x+1}{1-x}

Therefore:

\tan^{-1}\left(\frac{2x^2-x+1}{1-x}\right)=\tan^{-1}(-7)

Equate the arguments:

\frac{2x^2-x+1}{1-x}=-7

2x^2-x+1=-7+7x

2x^2-8x+8=0

x^2-4x+4=0

(x-2)^2=0

x=2

Answer:

x = 2

, i.e. Option C.

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