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Inverse Trigonometric Functions question

2024 · Shift 2 · Q18
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Inverse Trigonometric Functions question

2024 · Shift 2 · Q18

JEE AdvancedMathematicsInverse Trigonometric FunctionsMCQ+3 / −1
Considering only the principal values of the inverse trigonometric functions, the value of tan⁡(sin⁡−1(35)−2cos⁡−1(25))\tan \left(\sin ^{-1}\left(\frac{3}{5}\right)-2 \cos ^{-1}\left(\frac{2}{\sqrt{5}}\right)\right)tan(sin−1(53​)−2cos−1(5​2​)) is
  1. A
    724\frac{7}{24}247​
  2. B
    −724\frac{-7}{24}24−7​
  3. C
    −524\frac{-5}{24}24−5​
  4. D
    524\frac{5}{24}245​
View written solutionFree

Correct answer: B

  1. Let α=sin⁡−1(35),β=cos⁡−1(25).\alpha=\sin^{-1}\left(\frac35\right),\qquad \beta=\cos^{-1}\left(\frac{2}{\sqrt5}\right).α=sin−1(53​),β=cos−1(5​2​). We need to find tan⁡(α−2β).\tan(\alpha-2\beta).tan(α−2β).

  2. First, evaluate tan⁡α\tan\alphatanα.

Since α\alphaα is a principal value of sin⁡−1(35)\sin^{-1}\left(\frac35\right)sin−1(53​), we have sin⁡α=35,α∈[−π2,π2].\sin\alpha=\frac35,\qquad \alpha\in\left[-\frac\pi2,\frac\pi2\right].sinα=53​,α∈[−2π​,2π​]. So α\alphaα lies in the first quadrant, hence

\sqrt{1-\frac{9}{25}}=\frac45.$$ Therefore, $$\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{3/5}{4/5}=\frac34.$$ 3. Now evaluate $\tan\beta$. Since $\beta$ is a principal value of $\cos^{-1}\left(\frac{2}{\sqrt5}\right)$, we have $$\cos\beta=\frac{2}{\sqrt5},\qquad \beta\in[0,\pi].$$ Because $\cos\beta>0$, $\beta$ is in the first quadrant. Hence $$\sin\beta=\sqrt{1-\cos^2\beta}= \sqrt{1-\frac{4}{5}}=\frac{1}{\sqrt5}.$$ Thus, $$\tan\beta=\frac{\sin\beta}{\cos\beta}=\frac{1/\sqrt5}{2/\sqrt5}=\frac12.$$ 4. Use the double-angle identity for tangent: $$\tan 2\beta=\frac{2\tan\beta}{1-\tan^2\beta}.

Substituting tan⁡β=12\tan\beta=\frac12tanβ=21​,

=\frac{1}{1-\frac14} =\frac{1}{\frac34} =\frac43.$$ 5. Now use $$\tan(\alpha-2\beta)=\frac{\tan\alpha-\tan 2\beta}{1+\tan\alpha\tan 2\beta}.$$ So, $$\tan(\alpha-2\beta)= \frac{\frac34-\frac43}{1+\left(\frac34\right)\left(\frac43\right)}.

Compute numerator: 34−43=9−1612=−712.\frac34-\frac43=\frac{9-16}{12}= -\frac{7}{12}.43​−34​=129−16​=−127​. Compute denominator: 1+(34)(43)=1+1=2.1+\left(\frac34\right)\left(\frac43\right)=1+1=2.1+(43​)(34​)=1+1=2. Hence, tan⁡(α−2β)=−7122=−724.\tan(\alpha-2\beta)=\frac{-\frac{7}{12}}{2}=-\frac{7}{24}.tan(α−2β)=2−127​​=−247​.

  1. Therefore, the required value is −724.\boxed{-\frac{7}{24}}.−247​​.

  2. Checking options:

  • A: 724\frac{7}{24}247​
  • B: −724-\frac{7}{24}−247​
  • C: −524-\frac{5}{24}−245​
  • D: 524\frac{5}{24}245​

So the correct option is B.

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