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Inverse Trigonometric Functions question

2017 · 8 Apr · Shift 1 · Q35
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  5. /2017 · 8 Apr · Shift 1 · Q35

Inverse Trigonometric Functions question

2017 · 8 Apr · Shift 1 · Q35

JEE MainMathematicsInverse Trigonometric FunctionsMCQ+4 / −1
The value of tan-1 [1+x2+1−x21+x2−1−x2],∣x∣<12,xe0,\left[ {{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} } \over {\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right],\left| x \right| \lt {1 \over 2},x e 0,[1+x2​−1−x2​1+x2​+1−x2​​],∣x∣<21​,xe0, is equal to :
  1. A
    π4+12cos⁡−1 x2{\pi \over 4} + {1 \over 2}{\cos ^{ - 1}}\,{x^2}4π​+21​cos−1x2
  2. B
    π4+cos⁡−1 x2{\pi \over 4} + {\cos ^{ - 1}}\,{x^2}4π​+cos−1x2
  3. C
    π4−12cos⁡−1 x2{\pi \over 4} - {1 \over 2}{\cos ^{ - 1}}\,{x^2}4π​−21​cos−1x2
  4. D
    π4−cos⁡−1 x2{\pi \over 4} - {\cos ^{ - 1}}\,{x^2}4π​−cos−1x2
View written solutionFree

Correct answer: A

  1. Let A=1+x2,B=1−x2.A=\sqrt{1+x^2},\qquad B=\sqrt{1-x^2}.A=1+x2​,B=1−x2​. Then the given expression is tan⁡−1(A+BA−B).\tan^{-1}\left(\frac{A+B}{A-B}\right).tan−1(A−BA+B​).

  2. Simplify the ratio. Multiply numerator and denominator by A+BA+BA+B: A+BA−B⋅A+BA+B=(A+B)2A2−B2.\frac{A+B}{A-B}\cdot \frac{A+B}{A+B}=\frac{(A+B)^2}{A^2-B^2}.A−BA+B​⋅A+BA+B​=A2−B2(A+B)2​. Now, A2−B2=(1+x2)−(1−x2)=2x2,A^2-B^2=(1+x^2)-(1-x^2)=2x^2,A2−B2=(1+x2)−(1−x2)=2x2, and (A+B)2=A2+B2+2AB=(1+x2)+(1−x2)+2(1+x2)(1−x2)(A+B)^2=A^2+B^2+2AB=(1+x^2)+(1-x^2)+2\sqrt{(1+x^2)(1-x^2)}(A+B)2=A2+B2+2AB=(1+x2)+(1−x2)+2(1+x2)(1−x2)​ =2+21−x4=2(1+1−x4).=2+2\sqrt{1-x^4}=2(1+\sqrt{1-x^4}).=2+21−x4​=2(1+1−x4​). Therefore, A+BA−B=2(1+1−x4)2x2=1+1−x4x2.\frac{A+B}{A-B}=\frac{2(1+\sqrt{1-x^4})}{2x^2}=\frac{1+\sqrt{1-x^4}}{x^2}.A−BA+B​=2x22(1+1−x4​)​=x21+1−x4​​.

So the given value becomes tan⁡−1(1+1−x4x2).\tan^{-1}\left(\frac{1+\sqrt{1-x^4}}{x^2}\right).tan−1(x21+1−x4​​).

  1. Use the standard identity: If tan⁡θ=1+1−t2t,0<t≤1,\tan \theta=\frac{1+\sqrt{1-t^2}}{t}, \qquad 0<t\le 1,tanθ=t1+1−t2​​,0<t≤1, then θ=π4+12cos⁡−1t.\theta=\frac{\pi}{4}+\frac{1}{2}\cos^{-1} t.θ=4π​+21​cos−1t.

Here take t=x2.t=x^2.t=x2. Since ∣x∣<12|x|<\frac12∣x∣<21​ and x≠0x\ne 0x=0, we have 0<x2<14,0<x^2<\frac14,0<x2<41​, so this is valid.

Thus, tan⁡−1(1+1−x4x2)=π4+12cos⁡−1(x2).\tan^{-1}\left(\frac{1+\sqrt{1-x^4}}{x^2}\right)=\frac{\pi}{4}+\frac12\cos^{-1}(x^2).tan−1(x21+1−x4​​)=4π​+21​cos−1(x2).

  1. Verification of the identity. Let ϕ=cos⁡−1(x2),\phi=\cos^{-1}(x^2),ϕ=cos−1(x2), so that cos⁡ϕ=x2,sin⁡ϕ=1−x4.\cos\phi=x^2, \qquad \sin\phi=\sqrt{1-x^4}.cosϕ=x2,sinϕ=1−x4​. Now consider tan⁡(π4+ϕ2)=1+tan⁡(ϕ/2)1−tan⁡(ϕ/2).\tan\left(\frac\pi4+\frac\phi2\right)=\frac{1+\tan(\phi/2)}{1-\tan(\phi/2)}.tan(4π​+2ϕ​)=1−tan(ϕ/2)1+tan(ϕ/2)​. Using tan⁡ϕ2=sin⁡ϕ1+cos⁡ϕ=1−x41+x2,\tan\frac\phi2=\frac{\sin\phi}{1+\cos\phi}=\frac{\sqrt{1-x^4}}{1+x^2},tan2ϕ​=1+cosϕsinϕ​=1+x21−x4​​, we get
=\frac{1+x^2+\sqrt{1-x^4}}{1+x^2-\sqrt{1-x^4}}.$$ Multiply numerator and denominator by $1+x^2+\sqrt{1-x^4}$: $$=\frac{(1+x^2+\sqrt{1-x^4})^2}{(1+x^2)^2-(1-x^4)}.$$ The denominator is $$1+2x^2+x^4-1+x^4=2x^2(1+x^2).$$ This simplifies to $$\frac{1+\sqrt{1-x^4}}{x^2},$$ which matches our expression. Hence, $$\tan^{-1}\left(\frac{1+\sqrt{1-x^4}}{x^2}\right)=\frac\pi4+\frac12\cos^{-1}(x^2).$$ 5. Therefore the correct option is $$\boxed{\text{A }\left(\frac\pi4+\frac12\cos^{-1}x^2\right)}.$$
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