Q66 / 160 · TS EAMCET 2024 (10 May Shift 1) · Mathematics
If $\sin ^{-1}(4 x)-\cos ^{-1}(3 x)=\frac{\text { g }}{6}$, then $x=$
- A$\frac{\sqrt{3}}{2 \sqrt{7}}$
- B$\frac{\sqrt{3}}{4 \sqrt{7}}$
- Correct answerC$\frac{\sqrt{3}}{2 \sqrt{13}}$
- D$\frac{\sqrt{3}}{4 \sqrt{13}}$
Solution
$\sin ^{-1}(4 x)-\cos ^{-1}(3 x)=\frac{\pi}{6}$....(i)
Let, $A=\sin ^{-1}(4 x), B=\cos ^{-1}(3 x)$ $\sin A=4 x \Rightarrow \cos A=\sqrt{1-16 x^2}$ $\cos B=3 x \Rightarrow \sin B=\sqrt{1-9 x^2}$ $A-B=\frac{\pi}{6} \Rightarrow \sin (A-B)=\frac{1}{2} \quad[$ From (i)] $\sin A \cos B-\cos A \sin B=\frac{1}{2}$ $\Rightarrow 12 x^2-\sqrt{1-16 x^2} \times \sqrt{1-9 x^2}=\frac{1}{2}$ $\Rightarrow\left(12 x^2-\frac{1}{2}\right)^2=\left(1-16 x^2\right)\left(1-9 x^2\right)$ (On squaring both sides) On solving, we will get $x=\frac{\sqrt{3}}{2 \sqrt{13}}$.
Let, $A=\sin ^{-1}(4 x), B=\cos ^{-1}(3 x)$ $\sin A=4 x \Rightarrow \cos A=\sqrt{1-16 x^2}$ $\cos B=3 x \Rightarrow \sin B=\sqrt{1-9 x^2}$ $A-B=\frac{\pi}{6} \Rightarrow \sin (A-B)=\frac{1}{2} \quad[$ From (i)] $\sin A \cos B-\cos A \sin B=\frac{1}{2}$ $\Rightarrow 12 x^2-\sqrt{1-16 x^2} \times \sqrt{1-9 x^2}=\frac{1}{2}$ $\Rightarrow\left(12 x^2-\frac{1}{2}\right)^2=\left(1-16 x^2\right)\left(1-9 x^2\right)$ (On squaring both sides) On solving, we will get $x=\frac{\sqrt{3}}{2 \sqrt{13}}$.
Hence, the correct answer is option: 'C'.