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:
\[\begin{align}
& A.\dfrac{\sqrt{145}}{12} \\
& B.\dfrac{\sqrt{145}}{10} \\
& C.\dfrac{\sqrt{146}}{12} \\
& D.\dfrac{\sqrt{145}}{11} \\
\end{align}\]
Answer
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Hint: 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:
\[\begin{align}
& \Rightarrow \text{co}{{\text{s}}^{-1}}\left( \dfrac{2}{3x} \right)+\text{co}{{\text{s}}^{-1}}\left( \dfrac{3}{4x} \right)=\dfrac{\pi }{2} \\
& \Rightarrow \text{co}{{\text{s}}^{-1}}\left( \left( \dfrac{2}{3x} \right)\left( \dfrac{3}{4x} \right)-\sqrt{1-{{\left( \dfrac{2}{3x} \right)}^{2}}}\sqrt{1-{{\left( \dfrac{3}{4x} \right)}^{2}}} \right)=\dfrac{\pi }{2} \\
\end{align}\]
Solving further gives:
\[\begin{align}
& \Rightarrow \text{co}{{\text{s}}^{-1}}\left( \dfrac{1}{2{{x}^{2}}}-\sqrt{1-\dfrac{4}{9{{x}^{2}}}}\sqrt{1-\dfrac{9}{16{{x}^{2}}}} \right)=\dfrac{\pi }{2} \\
& \Rightarrow \text{co}{{\text{s}}^{-1}}\left( \dfrac{1}{2{{x}^{2}}}-\sqrt{\dfrac{9{{x}^{2}}-4}{9{{x}^{2}}}}\sqrt{\dfrac{16{{x}^{2}}-9}{16{{x}^{2}}}} \right)=\dfrac{\pi }{2} \\
\end{align}\]
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:
\[\begin{align}
& \Rightarrow \dfrac{1}{2{{x}^{2}}}-\sqrt{\dfrac{9{{x}^{2}}-4}{9{{x}^{2}}}}\sqrt{\dfrac{16{{x}^{2}}-9}{16{{x}^{2}}}}=0 \\
& \Rightarrow \dfrac{1}{2{{x}^{2}}}=\sqrt{\dfrac{9{{x}^{2}}-4}{9{{x}^{2}}}}\sqrt{\dfrac{16{{x}^{2}}-9}{16{{x}^{2}}}} \\
\end{align}\]
Now, we know that \[\sqrt{9{{x}^{2}}}=3x\text{ and }\sqrt{16{{x}^{2}}}=4x\]
Substituting this in above equation we get:
\[\begin{align}
& \Rightarrow \dfrac{1}{2{{x}^{2}}}=\dfrac{\sqrt{9{{x}^{2}}-4}\sqrt{16{{x}^{2}}-9}}{3x\times 4x} \\
& \Rightarrow \dfrac{1}{2{{x}^{2}}}=\dfrac{\sqrt{9{{x}^{2}}-4}\sqrt{16{{x}^{2}}-9}}{12{{x}^{2}}} \\
\end{align}\]
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:
\[\begin{align}
& 36=144{{x}^{4}}-81{{x}^{2}}-64{{x}^{2}}+36 \\
& \Rightarrow 36=144{{x}^{4}}-145{{x}^{2}}+36 \\
\end{align}\]
Cancelling 36 by subtracting 36 on both sides we get:
\[\begin{align}
& 0=144{{x}^{4}}-145{{x}^{2}} \\
& \Rightarrow 144{{x}^{4}}=145{{x}^{2}} \\
& \Rightarrow 144{{x}^{4}}-145{{x}^{2}}=0 \\
& \Rightarrow {{x}^{2}}\left( 144{{x}^{2}}-145 \right)=0 \\
\end{align}\]
So, we can equate the terms to 0 as
\[\begin{align}
& {{x}^{2}}=0\Rightarrow x=0\Rightarrow 144{{x}^{2}}-145=0 \\
& \Rightarrow {{x}^{2}}=\dfrac{145}{144}\Rightarrow x=\pm \sqrt{\dfrac{145}{144}}=\pm \dfrac{\sqrt{145}}{12} \\
\end{align}\]
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:
\[\begin{align}
& \Rightarrow {{\left( \dfrac{2}{3x} \right)}^{2}}={{\left( \sqrt{1-{{\left( \dfrac{3}{4x} \right)}^{2}}} \right)}^{2}} \\
& \Rightarrow \dfrac{4}{9{{x}^{2}}}=1-{{\left( \dfrac{3}{4x} \right)}^{2}} \\
& \Rightarrow \dfrac{4}{9{{x}^{2}}}+\dfrac{9}{16{{x}^{2}}}=1 \\
& \Rightarrow \dfrac{4}{9{{x}^{2}}}=1-\dfrac{9}{16{{x}^{2}}} \\
\end{align}\]
Taking LCM in RHS, we get:
\[\Rightarrow \dfrac{4}{9{{x}^{2}}}=\dfrac{16{{x}^{2}}-9}{16{{x}^{2}}}\]
Cross multiplying we get:
\[\begin{align}
& \Rightarrow 4\left( 16{{x}^{2}} \right)=\left( 9{{x}^{2}} \right)\left( 16{{x}^{2}}-9 \right) \\
& \Rightarrow 64{{x}^{2}}=144{{x}^{4}}-81{{x}^{2}} \\
& \Rightarrow 144{{x}^{4}}-145{{x}^{2}}=0 \\
& \Rightarrow {{x}^{2}}\left( 144{{x}^{2}}-145 \right)=0 \\
\end{align}\]
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.
\[\text{co}{{\text{s}}^{-1}}A+\text{co}{{\text{s}}^{-1}}B=\text{co}{{\text{s}}^{-1}}\left( AB-\sqrt{1-{{A}^{2}}}\sqrt{1-{{B}^{2}}} \right)\]
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:
\[\begin{align}
& \Rightarrow \text{co}{{\text{s}}^{-1}}\left( \dfrac{2}{3x} \right)+\text{co}{{\text{s}}^{-1}}\left( \dfrac{3}{4x} \right)=\dfrac{\pi }{2} \\
& \Rightarrow \text{co}{{\text{s}}^{-1}}\left( \left( \dfrac{2}{3x} \right)\left( \dfrac{3}{4x} \right)-\sqrt{1-{{\left( \dfrac{2}{3x} \right)}^{2}}}\sqrt{1-{{\left( \dfrac{3}{4x} \right)}^{2}}} \right)=\dfrac{\pi }{2} \\
\end{align}\]
Solving further gives:
\[\begin{align}
& \Rightarrow \text{co}{{\text{s}}^{-1}}\left( \dfrac{1}{2{{x}^{2}}}-\sqrt{1-\dfrac{4}{9{{x}^{2}}}}\sqrt{1-\dfrac{9}{16{{x}^{2}}}} \right)=\dfrac{\pi }{2} \\
& \Rightarrow \text{co}{{\text{s}}^{-1}}\left( \dfrac{1}{2{{x}^{2}}}-\sqrt{\dfrac{9{{x}^{2}}-4}{9{{x}^{2}}}}\sqrt{\dfrac{16{{x}^{2}}-9}{16{{x}^{2}}}} \right)=\dfrac{\pi }{2} \\
\end{align}\]
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:
\[\begin{align}
& \Rightarrow \dfrac{1}{2{{x}^{2}}}-\sqrt{\dfrac{9{{x}^{2}}-4}{9{{x}^{2}}}}\sqrt{\dfrac{16{{x}^{2}}-9}{16{{x}^{2}}}}=0 \\
& \Rightarrow \dfrac{1}{2{{x}^{2}}}=\sqrt{\dfrac{9{{x}^{2}}-4}{9{{x}^{2}}}}\sqrt{\dfrac{16{{x}^{2}}-9}{16{{x}^{2}}}} \\
\end{align}\]
Now, we know that \[\sqrt{9{{x}^{2}}}=3x\text{ and }\sqrt{16{{x}^{2}}}=4x\]
Substituting this in above equation we get:
\[\begin{align}
& \Rightarrow \dfrac{1}{2{{x}^{2}}}=\dfrac{\sqrt{9{{x}^{2}}-4}\sqrt{16{{x}^{2}}-9}}{3x\times 4x} \\
& \Rightarrow \dfrac{1}{2{{x}^{2}}}=\dfrac{\sqrt{9{{x}^{2}}-4}\sqrt{16{{x}^{2}}-9}}{12{{x}^{2}}} \\
\end{align}\]
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:
\[\begin{align}
& 36=144{{x}^{4}}-81{{x}^{2}}-64{{x}^{2}}+36 \\
& \Rightarrow 36=144{{x}^{4}}-145{{x}^{2}}+36 \\
\end{align}\]
Cancelling 36 by subtracting 36 on both sides we get:
\[\begin{align}
& 0=144{{x}^{4}}-145{{x}^{2}} \\
& \Rightarrow 144{{x}^{4}}=145{{x}^{2}} \\
& \Rightarrow 144{{x}^{4}}-145{{x}^{2}}=0 \\
& \Rightarrow {{x}^{2}}\left( 144{{x}^{2}}-145 \right)=0 \\
\end{align}\]
So, we can equate the terms to 0 as
\[\begin{align}
& {{x}^{2}}=0\Rightarrow x=0\Rightarrow 144{{x}^{2}}-145=0 \\
& \Rightarrow {{x}^{2}}=\dfrac{145}{144}\Rightarrow x=\pm \sqrt{\dfrac{145}{144}}=\pm \dfrac{\sqrt{145}}{12} \\
\end{align}\]
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:
\[\begin{align}
& \Rightarrow {{\left( \dfrac{2}{3x} \right)}^{2}}={{\left( \sqrt{1-{{\left( \dfrac{3}{4x} \right)}^{2}}} \right)}^{2}} \\
& \Rightarrow \dfrac{4}{9{{x}^{2}}}=1-{{\left( \dfrac{3}{4x} \right)}^{2}} \\
& \Rightarrow \dfrac{4}{9{{x}^{2}}}+\dfrac{9}{16{{x}^{2}}}=1 \\
& \Rightarrow \dfrac{4}{9{{x}^{2}}}=1-\dfrac{9}{16{{x}^{2}}} \\
\end{align}\]
Taking LCM in RHS, we get:
\[\Rightarrow \dfrac{4}{9{{x}^{2}}}=\dfrac{16{{x}^{2}}-9}{16{{x}^{2}}}\]
Cross multiplying we get:
\[\begin{align}
& \Rightarrow 4\left( 16{{x}^{2}} \right)=\left( 9{{x}^{2}} \right)\left( 16{{x}^{2}}-9 \right) \\
& \Rightarrow 64{{x}^{2}}=144{{x}^{4}}-81{{x}^{2}} \\
& \Rightarrow 144{{x}^{4}}-145{{x}^{2}}=0 \\
& \Rightarrow {{x}^{2}}\left( 144{{x}^{2}}-145 \right)=0 \\
\end{align}\]
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.
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