A possible value of \tan ( 1 \over 4\sin ^ - 1\sqrt 63 \over 8 ) is :

Inverse Trigonometric Functions PYQ — JEE Main 2021 · 24 Feb · Shift 2 · Q32 with Solution

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Correct answer: B

Let $\theta=\sin^{-1}\left(\frac{\sqrt{63}}{8}\right).$ Then $\sin\theta=\frac{\sqrt{63}}{8}=\frac{3\sqrt7}{8}.$ Since $\theta\in\left[-\frac\pi2,\frac\pi2\right]$ , here $\theta$ is acute.
Find $\cos\theta$ :

=\sqrt{1-\frac{63}{64}} =\sqrt{\frac1{64}} =\frac18.$$ So, $$\sin\theta=\frac{3\sqrt7}{8},\qquad \cos\theta=\frac18.$$ Hence, $$\tan\theta=\frac{\sin\theta}{\cos\theta}=3\sqrt7.$$ 3. We need $$\tan\left(\frac\theta4\right).$$ Let $$t=\tan\left(\frac\theta4\right).$$ Then $$\tan\theta=\tan(4\cdot \tfrac\theta4)=\tan(4\alpha)\quad \text{with }\alpha=\frac\theta4.$$ A more convenient route is to use repeated half-angle. First, compute $$\tan\frac\theta2=\frac{\sin\theta}{1+\cos\theta} =\frac{\frac{3\sqrt7}{8}}{1+\frac18} =\frac{\frac{3\sqrt7}{8}}{\frac98} =\frac{\sqrt7}{3}.$$ Now let $$u=\tan\frac\theta4.$$ Using $$\tan\frac{x}{2}=\frac{2\tan(x/4)}{1-\tan^2(x/4)}$$ with $x=\theta$, we get $$\frac{2u}{1-u^2}=\frac{\sqrt7}{3}.$$ 4. Solve for $u$: $$6u=\sqrt7(1-u^2)$$ $$\sqrt7u^2+6u-\sqrt7=0.$$ This quadratic gives $$u=\frac{-6\pm\sqrt{36+28}}{2\sqrt7} =\frac{-6\pm 8}{2\sqrt7}.$$ So, $$u=\frac{2}{2\sqrt7}=\frac1{\sqrt7} \quad \text{or} \quad u=\frac{-14}{2\sqrt7}=-\sqrt7.$$ 5. Since $\theta$ is acute and small enough, $\frac\theta4\in(0,\pi/2)$, so its tangent must be positive and less than $1$. Therefore, $$\tan\left(\frac14\sin^{-1}\frac{\sqrt{63}}8\right)=\frac1{\sqrt7}.$$ 6. Checking options: - A: $\sqrt7-1$ ✗ - B: $\dfrac1{\sqrt7}$ ✓ - C: $2\sqrt2-1$ ✗ - D: $\dfrac1{2\sqrt2}$ ✗ Therefore, the correct option is **B**.

View written solutionFree

Correct answer: B

Let $\theta=\sin^{-1}\left(\frac{\sqrt{63}}{8}\right).$ Then $\sin\theta=\frac{\sqrt{63}}{8}=\frac{3\sqrt7}{8}.$ Since $\theta\in\left[-\frac\pi2,\frac\pi2\right]$ , here $\theta$ is acute.
Find $\cos\theta$ :

=\sqrt{1-\frac{63}{64}} =\sqrt{\frac1{64}} =\frac18.$$ So, $$\sin\theta=\frac{3\sqrt7}{8},\qquad \cos\theta=\frac18.$$ Hence, $$\tan\theta=\frac{\sin\theta}{\cos\theta}=3\sqrt7.$$ 3. We need $$\tan\left(\frac\theta4\right).$$ Let $$t=\tan\left(\frac\theta4\right).$$ Then $$\tan\theta=\tan(4\cdot \tfrac\theta4)=\tan(4\alpha)\quad \text{with }\alpha=\frac\theta4.$$ A more convenient route is to use repeated half-angle. First, compute $$\tan\frac\theta2=\frac{\sin\theta}{1+\cos\theta} =\frac{\frac{3\sqrt7}{8}}{1+\frac18} =\frac{\frac{3\sqrt7}{8}}{\frac98} =\frac{\sqrt7}{3}.$$ Now let $$u=\tan\frac\theta4.$$ Using $$\tan\frac{x}{2}=\frac{2\tan(x/4)}{1-\tan^2(x/4)}$$ with $x=\theta$, we get $$\frac{2u}{1-u^2}=\frac{\sqrt7}{3}.$$ 4. Solve for $u$: $$6u=\sqrt7(1-u^2)$$ $$\sqrt7u^2+6u-\sqrt7=0.$$ This quadratic gives $$u=\frac{-6\pm\sqrt{36+28}}{2\sqrt7} =\frac{-6\pm 8}{2\sqrt7}.$$ So, $$u=\frac{2}{2\sqrt7}=\frac1{\sqrt7} \quad \text{or} \quad u=\frac{-14}{2\sqrt7}=-\sqrt7.$$ 5. Since $\theta$ is acute and small enough, $\frac\theta4\in(0,\pi/2)$, so its tangent must be positive and less than $1$. Therefore, $$\tan\left(\frac14\sin^{-1}\frac{\sqrt{63}}8\right)=\frac1{\sqrt7}.$$ 6. Checking options: - A: $\sqrt7-1$ ✗ - B: $\dfrac1{\sqrt7}$ ✓ - C: $2\sqrt2-1$ ✗ - D: $\dfrac1{2\sqrt2}$ ✗ Therefore, the correct option is **B**.