\displaystyle \textbf{Question 1: }\text{If }\cos^{-1}\frac{x}{2}+\cos^{-1}\frac{y}{3}=\alpha,\text{ then prove that}
\displaystyle 9x^2-12xy\cos\alpha+4y^2=36\sin^2\alpha
\displaystyle \text{Answer:}
\displaystyle \cos^{-1}\frac{x}{2}+\cos^{-1}\frac{y}{3}=\alpha
\displaystyle \text{Taking cosine on both sides,}
\displaystyle \cos\left(\cos^{-1}\frac{x}{2}+\cos^{-1}\frac{y}{3}\right)=\cos\alpha
\displaystyle \Rightarrow\frac{x}{2}\cdot\frac{y}{3}-\sqrt{1-\frac{x^2}{4}}\sqrt{1-\frac{y^2}{9}}=\cos\alpha
\displaystyle \Rightarrow\frac{xy}{6}-\frac{\sqrt{4-x^2}}{2}\cdot\frac{\sqrt{9-y^2}}{3}=\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 \text{Squaring both sides,}
\displaystyle (4-x^2)(9-y^2)=(xy-6\cos\alpha)^2
\displaystyle \Rightarrow36-4y^2-9x^2+x^2y^2=x^2y^2-12xy\cos\alpha+36\cos^2\alpha
\displaystyle \Rightarrow9x^2-12xy\cos\alpha+4y^2=36-36\cos^2\alpha
\displaystyle \Rightarrow9x^2-12xy\cos\alpha+4y^2=36\sin^2\alpha
\displaystyle \therefore 9x^2-12xy\cos\alpha+4y^2=36\sin^2\alpha
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Solve the equation: }\cos^{-1}\frac{a}{x}-\cos^{-1}\frac{b}{x}=\cos^{-1}\frac1b-\cos^{-1}\frac1a
\displaystyle \text{Answer:}
\displaystyle \cos^{-1}\frac{a}{x}-\cos^{-1}\frac{b}{x}=\cos^{-1}\frac1b-\cos^{-1}\frac1a
\displaystyle \Rightarrow\cos^{-1}\frac{a}{x}+\cos^{-1}\frac1a=\cos^{-1}\frac1b+\cos^{-1}\frac{b}{x}
\displaystyle \text{Using Property VIII,}
\displaystyle \cos^{-1}\left[\frac{a}{x}\cdot\frac1a-\sqrt{1-\frac{a^2}{x^2}}\sqrt{1-\frac1{a^2}}\right]
\displaystyle =\cos^{-1}\left[\frac{b}{x}\cdot\frac1b-\sqrt{1-\frac{b^2}{x^2}}\sqrt{1-\frac1{b^2}}\right]
\displaystyle \Rightarrow\frac1x-\sqrt{1-\frac{a^2}{x^2}}\sqrt{1-\frac1{a^2}}=\frac1x-\sqrt{1-\frac{b^2}{x^2}}\sqrt{1-\frac1{b^2}}
\displaystyle \Rightarrow\sqrt{1-\frac{a^2}{x^2}}\sqrt{1-\frac1{a^2}}=\sqrt{1-\frac{b^2}{x^2}}\sqrt{1-\frac1{b^2}}
\displaystyle \Rightarrow\left(1-\frac{a^2}{x^2}\right)\left(1-\frac1{a^2}\right)=\left(1-\frac{b^2}{x^2}\right)\left(1-\frac1{b^2}\right)
\displaystyle \Rightarrow1-\frac1{a^2}-\frac{a^2}{x^2}+\frac1{x^2}=1-\frac1{b^2}-\frac{b^2}{x^2}+\frac1{x^2}
\displaystyle \Rightarrow\frac{a^2-b^2}{x^2}=\frac1{b^2}-\frac1{a^2}
\displaystyle \Rightarrow\frac{a^2-b^2}{x^2}=\frac{a^2-b^2}{a^2b^2}
\displaystyle \Rightarrow x^2=a^2b^2
\displaystyle \Rightarrow x=\pm ab
\displaystyle \text{If }a,b>0\text{ and }x>0,\text{ then }x=ab.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Solve: }\cos^{-1}(\sqrt3x)+\cos^{-1}x=\frac{\pi}{2}
\displaystyle \text{Answer:}
\displaystyle \cos^{-1}(\sqrt3x)+\cos^{-1}x=\frac{\pi}{2}
\displaystyle \text{Taking cosine on both sides,}
\displaystyle \cos\left[\cos^{-1}(\sqrt3x)+\cos^{-1}x\right]=\cos\frac{\pi}{2}
\displaystyle \Rightarrow\sqrt3x^2-\sqrt{1-3x^2}\sqrt{1-x^2}=0
\displaystyle \Rightarrow\sqrt3x^2=\sqrt{1-3x^2}\sqrt{1-x^2}
\displaystyle \text{Squaring both sides,}
\displaystyle 3x^4=(1-3x^2)(1-x^2)
\displaystyle \Rightarrow3x^4=1-x^2-3x^2+3x^4
\displaystyle \Rightarrow4x^2=1
\displaystyle \Rightarrow x^2=\frac14
\displaystyle \Rightarrow x=\pm\frac12
\displaystyle \text{For }x=\frac12,
\displaystyle \cos^{-1}\left(\frac{\sqrt3}{2}\right)+\cos^{-1}\left(\frac12\right)=\frac{\pi}{6}+\frac{\pi}{3}=\frac{\pi}{2}.
\displaystyle \text{For }x=-\frac12,
\displaystyle \cos^{-1}\left(-\frac{\sqrt3}{2}\right)+\cos^{-1}\left(-\frac12\right)=\frac{5\pi}{6}+\frac{2\pi}{3}=\frac{3\pi}{2}\neq\frac{\pi}{2}.
\displaystyle \therefore x=\frac12
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Prove that }\cos^{-1}\frac45+\cos^{-1}\frac{12}{13}=\cos^{-1}\frac{33}{65}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\cos^{-1}\frac45+\cos^{-1}\frac{12}{13}
\displaystyle \text{Since }\frac45+\frac{12}{13}>0,\text{ using Property VIII,}
\displaystyle \text{LHS}=\cos^{-1}\left[\frac45\cdot\frac{12}{13}-\sqrt{1-\left(\frac45\right)^2}\sqrt{1-\left(\frac{12}{13}\right)^2}\right]
\displaystyle =\cos^{-1}\left[\frac{48}{65}-\frac35\cdot\frac5{13}\right]
\displaystyle =\cos^{-1}\left(\frac{48}{65}-\frac{15}{65}\right)
\displaystyle =\cos^{-1}\frac{33}{65}
\displaystyle =\text{RHS}
\displaystyle \\

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