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Inverse Trigonometric Functions question

2019 · 12 Apr · Shift 1 · Q32
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Inverse Trigonometric Functions question

2019 · 12 Apr · Shift 1 · Q32

JEE MainMathematicsInverse Trigonometric FunctionsMCQ+4 / −1
The value of sin⁡−1(1213)−sin⁡−1(35){\sin ^{ - 1}}\left( {{{12} \over {13}}} \right) - {\sin ^{ - 1}}\left( {{3 \over 5}} \right)sin−1(1312​)−sin−1(53​) is equal to :
  1. A
    π−sin⁡−1(6365)\pi - {\sin ^{ - 1}}\left( {{{63} \over {65}}} \right)π−sin−1(6563​)
  2. B
    π2−sin⁡−1(5665){\pi \over 2} - {\sin ^{ - 1}}\left( {{{56} \over {65}}} \right)2π​−sin−1(6556​)
  3. C
    π2−cos⁡−1(965){\pi \over 2} - {\cos ^{ - 1}}\left( {{9 \over {65}}} \right)2π​−cos−1(659​)
  4. D
    π−cos⁡−1(3365)\pi - {\cos ^{ - 1}}\left( {{{33} \over {65}}} \right)π−cos−1(6533​)
View written solutionFree

Correct answer: B

  1. Let α=sin⁡−1(1213),β=sin⁡−1(35).\alpha=\sin^{-1}\left(\frac{12}{13}\right),\qquad \beta=\sin^{-1}\left(\frac35\right).α=sin−1(1312​),β=sin−1(53​). Then we need to find α−β.\alpha-\beta.α−β.

  2. Since both angles are principal values of inverse sine, we have α,β∈[−π2,π2].\alpha,\beta\in\left[-\frac\pi2,\frac\pi2\right].α,β∈[−2π​,2π​]. Also, sin⁡α=1213,sin⁡β=35.\sin\alpha=\frac{12}{13},\qquad \sin\beta=\frac35.sinα=1312​,sinβ=53​. Because both are positive, both angles are in the first quadrant.

So, cos⁡α=1−(1213)2=513,\cos\alpha=\sqrt{1-\left(\frac{12}{13}\right)^2}=\frac5{13},cosα=1−(1312​)2​=135​, cos⁡β=1−(35)2=45.\cos\beta=\sqrt{1-\left(\frac35\right)^2}=\frac45.cosβ=1−(53​)2​=54​.

  1. Now use sin⁡(α−β)=sin⁡αcos⁡β−cos⁡αsin⁡β.\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta.sin(α−β)=sinαcosβ−cosαsinβ. Thus,
=\frac{48}{65}-\frac{15}{65}=\frac{33}{65}.$$ 4. Also, $$\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta$$ $$=\frac5{13}\cdot\frac45+\frac{12}{13}\cdot\frac35 =\frac{20}{65}+\frac{36}{65}=\frac{56}{65}.$$ Since both sine and cosine are positive, $\alpha-\beta$ lies in the first quadrant. Hence, $$\alpha-\beta=\sin^{-1}\left(\frac{33}{65}\right)=\cos^{-1}\left(\frac{56}{65}\right).$$ 5. Now compare with the options. Using the identity $$\sin^{-1}x+\cos^{-1}x=\frac\pi2,$$ we get $$\cos^{-1}\left(\frac{56}{65}\right)=\frac\pi2-\sin^{-1}\left(\frac{56}{65}\right).$$ Thus, $$\alpha-\beta=\frac\pi2-\sin^{-1}\left(\frac{56}{65}\right).$$ This matches **Option B**. 6. Quick check of other options: - **A:** $\pi-\sin^{-1}(63/65)$ is large, not in first quadrant. - **C:** $\frac\pi2-\cos^{-1}(9/65)=\sin^{-1}(9/65)\ne \sin^{-1}(33/65)$. - **D:** $\pi-\cos^{-1}(33/65)$ is also too large. Therefore, the correct answer is **B**.
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