Question

If the value of $\text{co}{{\text{s}}^{-1}}\left( \dfrac{2}{3x} \right)+\text{co}{{\text{s}}^{-1}}\left( \dfrac{3}{4x} \right)=\dfrac{\pi }{2}$ where $\left( x>\dfrac{3}{4} \right)$ then x equal to:

& A.\dfrac{\sqrt{145}}{12} \\\ & B.\dfrac{\sqrt{145}}{10} \\\ & C.\dfrac{\sqrt{146}}{12} \\\ & D.\dfrac{\sqrt{145}}{11} \\\ \end{aligned}$$

Answer 1

To solve this question, we will first try to simplify ${{\cos }^{-1}}$ on the LHS of equation by using formula:
$\text{co}{{\text{s}}^{-1}}A+\text{co}{{\text{s}}^{-1}}B=\text{co}{{\text{s}}^{-1}}\left( AB-\sqrt{1-{{A}^{2}}}\sqrt{1-{{B}^{2}}} \right)$
Where A and B are angles. Then we will take cos on both sides, use the identity $\text{cos}\left( \text{co}{{\text{s}}^{-1}}x \right)=x$ and result $\text{cos}\dfrac{\pi }{2}=0$ to simplify further. We will obtain an equation of degree 4 and get values of x. Check which values fall in the range of $\left( x>\dfrac{3}{4} \right)$ and report the answer.

Complete step-by-step solution:
We have a trigonometric identity stating relation between $\text{co}{{\text{s}}^{-1}}A+\text{co}{{\text{s}}^{-1}}B$ which is given as below:
$\text{co}{{\text{s}}^{-1}}A+\text{co}{{\text{s}}^{-1}}B=\text{co}{{\text{s}}^{-1}}\left( AB-\sqrt{1-{{A}^{2}}}\sqrt{1-{{B}^{2}}} \right)$
Where A and B are angles.
We are given that, $\text{co}{{\text{s}}^{-1}}\left( \dfrac{2}{3x} \right)+\text{co}{{\text{s}}^{-1}}\left( \dfrac{3}{4x} \right)=\dfrac{\pi }{2}$ where $\left( x>\dfrac{3}{4} \right)$
Using the above stated formula by substituting $A=\dfrac{2}{3x}\text{ and }B=\dfrac{3}{4x}$ in above we get:

& \Rightarrow \text{co}{{\text{s}}^{-1}}\left( \dfrac{2}{3x} \right)+\text{co}{{\text{s}}^{-1}}\left( \dfrac{3}{4x} \right)=\dfrac{\pi }{2} \\\ & \Rightarrow \text{co}{{\text{s}}^{-1}}\left( \left( \dfrac{2}{3x} \right)\left( \dfrac{3}{4x} \right)-\sqrt{1-{{\left( \dfrac{2}{3x} \right)}^{2}}}\sqrt{1-{{\left( \dfrac{3}{4x} \right)}^{2}}} \right)=\dfrac{\pi }{2} \\\ \end{aligned}$$ Solving further gives: $$\begin{aligned} & \Rightarrow \text{co}{{\text{s}}^{-1}}\left( \dfrac{1}{2{{x}^{2}}}-\sqrt{1-\dfrac{4}{9{{x}^{2}}}}\sqrt{1-\dfrac{9}{16{{x}^{2}}}} \right)=\dfrac{\pi }{2} \\\ & \Rightarrow \text{co}{{\text{s}}^{-1}}\left( \dfrac{1}{2{{x}^{2}}}-\sqrt{\dfrac{9{{x}^{2}}-4}{9{{x}^{2}}}}\sqrt{\dfrac{16{{x}^{2}}-9}{16{{x}^{2}}}} \right)=\dfrac{\pi }{2} \\\ \end{aligned}$$ Taking cos on both sides, we get: $$\Rightarrow \cos \left[ \text{co}{{\text{s}}^{-1}} \left( \dfrac{1}{2{{x}^{2}}}-\sqrt{\dfrac{9{{x}^{2}}-4}{9{{x}^{2}}}}\sqrt{\dfrac{16{{x}^{2}}-9}{16{{x}^{2}}}} \right)\right]=\cos \dfrac{\pi }{2}$$ Now, we know that $\cos \dfrac{\pi }{2}=0$ and also the identity $\text{cos}\left( \text{co}{{\text{s}}^{-1}}x \right)=x$ so, applying this we get: $$\begin{aligned} & \Rightarrow \dfrac{1}{2{{x}^{2}}}-\sqrt{\dfrac{9{{x}^{2}}-4}{9{{x}^{2}}}}\sqrt{\dfrac{16{{x}^{2}}-9}{16{{x}^{2}}}}=0 \\\ & \Rightarrow \dfrac{1}{2{{x}^{2}}}=\sqrt{\dfrac{9{{x}^{2}}-4}{9{{x}^{2}}}}\sqrt{\dfrac{16{{x}^{2}}-9}{16{{x}^{2}}}} \\\ \end{aligned}$$ Now, we know that $$\sqrt{9{{x}^{2}}}=3x\text{ and }\sqrt{16{{x}^{2}}}=4x$$ Substituting this in above equation we get: $$\begin{aligned} & \Rightarrow \dfrac{1}{2{{x}^{2}}}=\dfrac{\sqrt{9{{x}^{2}}-4}\sqrt{16{{x}^{2}}-9}}{3x\times 4x} \\\ & \Rightarrow \dfrac{1}{2{{x}^{2}}}=\dfrac{\sqrt{9{{x}^{2}}-4}\sqrt{16{{x}^{2}}-9}}{12{{x}^{2}}} \\\ \end{aligned}$$ Cancelling $\dfrac{1}{2{{x}^{2}}}$ from both sides we get: $$\Rightarrow 1=\dfrac{\sqrt{9{{x}^{2}}-4}\sqrt{16{{x}^{2}}-9}}{6}$$ Cross multiplying we get: $$\Rightarrow 6=\sqrt{9{{x}^{2}}-4}\sqrt{16{{x}^{2}}-9}$$ Squaring both sides, we get: $$36=\left( 9{{x}^{2}}-4 \right)\left( 16{{x}^{2}}-9 \right)$$ Solving the expression by opening bracket of RHS we get: $$\begin{aligned} & 36=144{{x}^{4}}-81{{x}^{2}}-64{{x}^{2}}+36 \\\ & \Rightarrow 36=144{{x}^{4}}-145{{x}^{2}}+36 \\\ \end{aligned}$$ Cancelling 36 by subtracting 36 on both sides we get: $$\begin{aligned} & 0=144{{x}^{4}}-145{{x}^{2}} \\\ & \Rightarrow 144{{x}^{4}}=145{{x}^{2}} \\\ & \Rightarrow 144{{x}^{4}}-145{{x}^{2}}=0 \\\ & \Rightarrow {{x}^{2}}\left( 144{{x}^{2}}-145 \right)=0 \\\ \end{aligned}$$ So, we can equate the terms to 0 as $$\begin{aligned} & {{x}^{2}}=0\Rightarrow x=0\Rightarrow 144{{x}^{2}}-145=0 \\\ & \Rightarrow {{x}^{2}}=\dfrac{145}{144}\Rightarrow x=\pm \sqrt{\dfrac{145}{144}}=\pm \dfrac{\sqrt{145}}{12} \\\ \end{aligned}$$ So the possible value of x are $$x=0,+\dfrac{\sqrt{145}}{12},-\dfrac{\sqrt{145}}{12}$$ Now, given that $\left( x>\dfrac{3}{4} \right)\Rightarrow x=0$ is not possible as $$\left( x>\dfrac{3}{4} \right)\Rightarrow x\text{ is positive }\Rightarrow \text{ x }=\text{ }-\dfrac{\sqrt{145}}{12}\text{ not possible}\text{.}$$ $$\Rightarrow \text{x}=\dfrac{\sqrt{145}}{12}$$ is the only possible value of x. Therefore, we have value of $$\text{x}=\dfrac{\sqrt{145}}{12}$$ so option A is correct. **Note:** Another way to solve this question is by using the trigonometric identity given as: $$\dfrac{\pi }{2}-{{\cos }^{-1}}\theta ={{\sin }^{-1}}\theta \text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)}$$ We have $$\text{co}{{\text{s}}^{-1}}\left( \dfrac{2}{3x} \right)+\text{co}{{\text{s}}^{-1}}\left( \dfrac{3}{4x} \right)=\dfrac{\pi }{2}$$ where $\left( x>\dfrac{3}{4} \right)$ Taking $$\text{co}{{\text{s}}^{-1}}\left( \dfrac{3}{4x} \right)$$ to other side we have $$\text{co}{{\text{s}}^{-1}}\left( \dfrac{2}{3x} \right)=\dfrac{\pi }{2}-\text{co}{{\text{s}}^{-1}}\left( \dfrac{3}{4x} \right)$$ Using the identity stated above in equation (i) by substituting $\theta =\dfrac{3}{4x}$ we get: $$\text{co}{{\text{s}}^{-1}}\left( \dfrac{2}{3x} \right)={{\sin }^{-1}}\left( \dfrac{3}{4x} \right)$$ Again we will use a trigonometric identity given as: $${{\sin }^{-1}}\theta ={{\cos }^{-1}}\sqrt{1-{{\theta }^{2}}}$$ Using this identity above by putting $\theta =\dfrac{3}{4x}$ we get: $$\text{co}{{\text{s}}^{-1}}\left( \dfrac{2}{3x} \right)=\text{co}{{\text{s}}^{-1}}\sqrt{1-{{\left( \dfrac{3}{4x} \right)}^{2}}}$$ Now, applying cos both sides and using $${{\cos }^{-1}}\cos \theta =\theta $$ we get: $$\Rightarrow \left( \dfrac{2}{3x} \right)=\sqrt{1-{{\left( \dfrac{3}{4x} \right)}^{2}}}$$ Squaring both sides, we get: $$\begin{aligned} & \Rightarrow {{\left( \dfrac{2}{3x} \right)}^{2}}={{\left( \sqrt{1-{{\left( \dfrac{3}{4x} \right)}^{2}}} \right)}^{2}} \\\ & \Rightarrow \dfrac{4}{9{{x}^{2}}}=1-{{\left( \dfrac{3}{4x} \right)}^{2}} \\\ & \Rightarrow \dfrac{4}{9{{x}^{2}}}+\dfrac{9}{16{{x}^{2}}}=1 \\\ & \Rightarrow \dfrac{4}{9{{x}^{2}}}=1-\dfrac{9}{16{{x}^{2}}} \\\ \end{aligned}$$ Taking LCM in RHS, we get: $$\Rightarrow \dfrac{4}{9{{x}^{2}}}=\dfrac{16{{x}^{2}}-9}{16{{x}^{2}}}$$ Cross multiplying we get: $$\begin{aligned} & \Rightarrow 4\left( 16{{x}^{2}} \right)=\left( 9{{x}^{2}} \right)\left( 16{{x}^{2}}-9 \right) \\\ & \Rightarrow 64{{x}^{2}}=144{{x}^{4}}-81{{x}^{2}} \\\ & \Rightarrow 144{{x}^{4}}-145{{x}^{2}}=0 \\\ & \Rightarrow {{x}^{2}}\left( 144{{x}^{2}}-145 \right)=0 \\\ \end{aligned}$$ So, we again have obtained $x=0,\pm \dfrac{\sqrt{145}}{12}$ as we have obtained above. Here, $$x>\dfrac{3}{4}\Rightarrow x=+\dfrac{\sqrt{145}}{12}$$ is only possible answer.