\displaystyle \text{Question 1: If:} \cos^{-1} \frac{x}{2} + \cos^{-1} \frac{y}{3} =  \alpha. \text{ then prove that } \\ \\  9x^2- 12xy \cos \alpha + 4y^2 = 36 \sin^2 \alpha       

\displaystyle  \text{Answer:}  

\displaystyle  \text{We know, }  \cos^{-1} x + \cos^{-1} y = \cos^{-1} [ xy - \sqrt{1-x^2} \sqrt{1-y^2}]

\displaystyle \text{Given, } \cos^{-1} \frac{x}{2} + \cos^{-1} \frac{y}{3} =  \alpha

\displaystyle \Rightarrow \cos^{-1} \Bigg[ \frac{x}{2} \cdot \frac{y}{3} - \sqrt{1 - \frac{x^2}{4}}  \sqrt{1 - \frac{y^2}{4}}  \Bigg]=  \alpha

\displaystyle \Rightarrow \frac{x}{2} \cdot \frac{y}{3} - \sqrt{1 - \frac{x^2}{4}}  \sqrt{1 - \frac{y^2}{4}} = \cos \alpha

\displaystyle \Rightarrow  xy - \sqrt{4-x^2} \sqrt{9-y^2} = 6 \cos \alpha

\displaystyle \Rightarrow \sqrt{4-x^2} \sqrt{9-y^2} = xy - 6 \cos \alpha

\displaystyle \Rightarrow  (4-x^2) ( 9-y^2)  = x^2 y^2 + 36 \cos^2 \alpha - 12 xy \cos \alpha

\displaystyle \Rightarrow  36 - 4y^2-9x^2+x^2y^2= x^2 y^2 + 36 \cos^2 \alpha - 12 xy \cos \alpha

\displaystyle \Rightarrow 9x^2 - 12 xy \cos \alpha+ 4y^2 = 36 - 36\cos^2 \alpha

\displaystyle \Rightarrow 9x^2 - 12 xy \cos \alpha+ 4y^2 = 36 \sin^2 \alpha

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\displaystyle \text{Question 2: Solve the equation:} \cos^{-1} \frac{a}{x} - \cos^{-1} \frac{b}{x} = \cos^{-1} \frac{1}{b} - \cos^{-1} \frac{1}{a}      

\displaystyle  \text{Answer:}  

\displaystyle \text{Given, } \cos^{-1} \frac{a}{x} - \cos^{-1} \frac{b}{x} = \cos^{-1} \frac{1}{b} - \cos^{-1} \frac{1}{a}

\displaystyle \Rightarrow \cos^{-1} \frac{a}{x} + \cos^{-1} \frac{1}{a}  = \cos^{-1} \frac{1}{b} + \cos^{-1} \frac{b}{x}

\displaystyle \Rightarrow \cos^{-1} \Bigg[ \frac{a}{x} \cdot  \frac{1}{a} - \sqrt{1 - \Big(\frac{a}{x}\Big)^2}  \sqrt{1 -\Big (\frac{1}{a}\Big)^2 }      \Bigg] = \cos^{-1} \Bigg[ \frac{b}{x} \cdot  \frac{1}{b} - \sqrt{1 - \Big(\frac{b}{x}\Big)^2}  \sqrt{1 - \Big(\frac{1}{b}\Big)^2 }      \Bigg]

\displaystyle  \text{We know, }  \cos^{-1} x + \cos^{-1} y = \cos^{-1} [ xy - \sqrt{1-x^2} \sqrt{1-y^2}]

\displaystyle \Rightarrow \cos^{-1} \Bigg[ \frac{1}{x}  - \sqrt{1 - \Big(\frac{a^2}{x^2}\Big)}  \sqrt{1 -\Big (\frac{1}{a^2}\Big) }      \Bigg] = \cos^{-1} \Bigg[ \frac{1}{x}  - \sqrt{1 - \Big(\frac{b^2}{x^2}\Big)}  \sqrt{1 - \Big(\frac{1}{b^2}\Big) }      \Bigg]

\displaystyle \Rightarrow  \frac{1}{x}  - \sqrt{1 - \Big(\frac{a^2}{x^2}\Big)}  \sqrt{1 -\Big (\frac{1}{a^2}\Big) }  =  \frac{1}{x}  - \sqrt{1 - \Big(\frac{b^2}{x^2}\Big)}  \sqrt{1 - \Big(\frac{1}{b^2}\Big) } 

\displaystyle \Rightarrow  \sqrt{1 - \Big(\frac{a^2}{x^2}\Big)}  \sqrt{1 -\Big (\frac{1}{a^2}\Big) }  =  \sqrt{1 - \Big(\frac{b^2}{x^2}\Big)}  \sqrt{1 - \Big(\frac{1}{b^2}\Big) } 

\displaystyle \Rightarrow \Big(1 - \frac{a^2}{x^2}\Big)\Big(1 - \frac{1}{a^2}\Big) = \Big(1 - \frac{b^2}{x^2}\Big)\Big(1 - \frac{1}{b^2}\Big)

\displaystyle \Rightarrow 1 - \frac{1}{a^2} - \frac{a^2}{x^2} + \frac{1}{x^2} = 1 - \frac{1}{b^2} - \frac{b^2}{x^2} + \frac{1}{x^2}

\displaystyle \Rightarrow \frac{a^2 - b^2}{x^2} = \frac{1}{b^2} - \frac{1}{a^2}

\displaystyle \Rightarrow  \frac{a^2 - b^2}{x^2} = \frac{a^2 - b^2 }{a^2b^2}

\displaystyle \Rightarrow  x^2 = a^2 b^2

\displaystyle \Rightarrow  x= ab

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\displaystyle \text{Question 3: Solve:} \cos^{-1}\sqrt{3} x + \cos^{-1} x = \frac{\pi}{2}      

\displaystyle  \text{Answer:}  

\displaystyle \text{Given, }  \cos^{-1}\sqrt{3} x + \cos^{-1} x = \frac{\pi}{2}

\displaystyle \Rightarrow  \cos^{-1} [ \sqrt{3} x \times x - \sqrt{ 1 - (\sqrt{3} x)^2  } \sqrt{ 1 - x^2  } ] = \frac{\pi}{2}

\displaystyle \Rightarrow   \sqrt{3} x \times x - \sqrt{ 1 - 3x^2  } \sqrt{ 1 - x^2  } = \cos \frac{\pi}{2}

\displaystyle \Rightarrow \sqrt{3} x \times x = \sqrt{ 1 - 3x^2  } \sqrt{ 1 - x^2  }

\displaystyle \Rightarrow  3x^4 = (1-3x^2)(1-x^2)

\displaystyle \Rightarrow  3x^4 = 1 - 3x^2 + 3x^4 - x^2

\displaystyle \Rightarrow 4x^2 = 1

\displaystyle \Rightarrow x^2 = \frac{1}{4}

\displaystyle \Rightarrow x = \pm \frac{1}{2}

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\displaystyle \text{Question 4: Prove that:} \cos^{-1} \frac{4}{5} + \cos^{-1} \frac{12}{13} =  \cos^{-1} \frac{33}{65}      

\displaystyle  \text{Answer:}  

\displaystyle \text{LHS  }= \cos^{-1} \frac{4}{5} + \cos^{-1} \frac{12}{13}

\displaystyle = \cos^{-1} \Bigg[  \frac{4}{5} \times  \frac{12}{13 } - \sqrt{1 - \Big( \frac{4}{5} \Big)^2}   \sqrt{1 - \Big( \frac{12}{13} \Big)^2}  \Bigg]

\displaystyle = \cos^{-1} \Big[ \frac{48}{65} - \frac{15}{65} \Big]

\displaystyle = \cos^{-1} \frac{33}{65} = \text{LHS  }

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