Question Statement

Solve the following inverse trigonometric equations algebraically and prove the given identities:

Q3. Solve for $x$: (i) ${\sin }^{-1}(x-1)=\frac{\pi }{2}$ (ii) ${\sin }^{-1}x={\cos }^{-1}\left(\frac{5}{13}\right)$ (iii) ${\tan }^{-1}x={\sin }^{-1}\frac{24}{25}$ (iv) ${\cos }^{-1}\left(x-\frac{\sqrt{3}}{2}\right)=\frac{\pi }{6}$ (v) ${\tan }^{-1}\left(x+\frac{\sqrt{2}}{2}\right)=\frac{\pi }{4}$ (vi) ${\cos }^{-1}\left(x-\frac{1}{2}\right)=\frac{\pi }{3}$ (vii) $6\mathrm{Arc}\cot (2x)-5\pi =0$ (viii) $9\mathrm{Arc}{\sin }^2(x)-{\pi }^2=0$ (ix) $\pi -2{\sin }^{-1}x=2\pi$ (x) $4{\mathrm{Arctan}}^2(x)-3\pi \mathrm{Arc}\tan x-{\pi }^2=0$ (xi) ${\tan }^{-1}\left(\frac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}}\right)=\frac{\pi }{4}-\frac{x}{2}$

Q4. Find the values of $x$: (i) ${\sin }^{-1}\left(\frac{\sqrt{3}}{2}\right)+{\cos }^{-1}x=\frac{\pi }{2}$ (ii) ${\sin }^{-1}\left(\frac{1}{2}\right)+{\cos }^{-1}(2x)=\frac{\pi }{6}$ (iii) ${\sin }^{-1}(x)+{\cos }^{-1}\left(\frac{4}{5}\right)=\pi$ (iv) ${\tan }^{-1}x-{\tan }^{-1}\frac{1}{x}=\frac{\pi }{2}$

Q5. Prove the following identities: (i) ${\sin }^{-1}\left(\frac{5}{13}\right)-{\sin }^{-1}\left(\frac{63}{65}\right)={\cos }^{-1}\left(\frac{3}{5}\right)$ (ii) ${\sin }^{-1}\left(\frac{8}{17}\right)+{\sin }^{-1}\left(\frac{3}{5}\right)={\tan }^{-1}\left(\frac{77}{36}\right)$ (iii) ${\sin }^{-1}\frac{3}{5}+{\cos }^{-1}\frac{12}{13}={\sin }^{-1}\left(\frac{56}{65}\right)$ (iv) ${\tan }^{-1}\left(\frac{1}{4}\right)+{\tan }^{-1}\left(\frac{2}{9}\right)=\frac{1}{2}{\tan }^{-1}\left(\frac{4}{3}\right)$ (v) ${\cos }^{-1}\left(\frac{4}{5}\right)+{\cos }^{-1}\left(\frac{12}{13}\right)={\cos }^{-1}\left(\frac{33}{65}\right)$ (vi) $2{\tan }^{-1}\left(\frac{3}{4}\right)-{\tan }^{-1}\left(\frac{17}{31}\right)=\frac{\pi }{4}$ (vii) ${\tan }^{-1}\left(\frac{1}{2}\right)+{\tan }^{-1}\left(\frac{1}{5}\right)+{\tan }^{-1}\left(\frac{1}{8}\right)=\frac{\pi }{4}$ (viii) ${\sin }^{-1}\left(\frac{4}{5}\right)+{\sin }^{-1}\left(\frac{5}{13}\right)+{\sin }^{-1}\left(\frac{16}{65}\right)=\frac{\pi }{2}$ (ix) ${\tan }^{-1}\left(\frac{3}{4}\right)+{\tan }^{-1}\left(\frac{3}{5}\right)-{\tan }^{-1}\left(\frac{8}{19}\right)=\frac{\pi }{4}$ (x) $2{\tan }^{-1}\left(\frac{1}{5}\right)+{\sec }^{-1}\left(\frac{5\sqrt{2}}{7}\right)+2{\tan }^{-1}\left(\frac{1}{8}\right)=\frac{\pi }{4}$ (xi) ${\sin }^{-1}(x)+{\cos }^{-1}(x)=\frac{\pi }{2}$ where $-1<x<1$

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Background and Explanation

To solve inverse trigonometric equations, we use the property that if $f^{-1}(x)=y$, then $f(y)=x$ within the appropriate domain and range. We often use right-triangle trigonometry (Pythagoras theorem) to convert between different inverse functions (e.g., converting ${\cos }^{-1}$ to ${\sin }^{-1}$). Additionally, sum and difference formulas for ${\sin }^{-1}$, ${\cos }^{-1}$, and ${\tan }^{-1}$ are essential for proving identities.

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Solution

Q3. Inverse Trigonometric Equations

Part (i)

${\sin }^{-1}(x-1)=\frac{\pi }{2}$ Taking $\sin$ on both sides: $\begin{array}{rl}\sin \left\{{\sin }^{-1}(x-1)\right\}& =\sin \frac{\pi }{2}\\ x-1& =1\\ x& =1+1\\ x& =2\end{array}$

Part (ii)

${\sin }^{-1}x={\cos }^{-1}\left(\frac{5}{13}\right)$ Let $\theta ={\cos }^{-1}\frac{5}{13}$, so $\cos \theta =\frac{5}{13}=\frac{Base}{Hypotenuse}$. Using Pythagoras theorem ($H^2=B^2+P^2$): $\begin{array}{rl}13^2& =5^2+p^2\\ 169& =25+p^2\\ p^2& =144⟹p=12\end{array}$ So, $\sin \theta =\frac{p}{H}=\frac{12}{13}⟹\theta ={\sin }^{-1}\frac{12}{13}$. Substituting back: $\begin{array}{rl}{\sin }^{-1}x& ={\sin }^{-1}\frac{12}{13}\\ x& =\frac{12}{13}\end{array}$

Part (iii)

${\tan }^{-1}x={\sin }^{-1}\frac{24}{25}$ Let $\theta ={\sin }^{-1}\frac{24}{25}$, so $\sin \theta =\frac{24}{25}=\frac{P}{H}$. Using Pythagoras theorem: $\begin{array}{rl}25^2& =B^2+24^2\\ 625& =B^2+576\\ B^2& =49⟹B=7\end{array}$ Then $\tan \theta =\frac{P}{B}=\frac{24}{7}⟹\theta ={\tan }^{-1}\frac{24}{7}$. $\begin{array}{rl}{\tan }^{-1}x& ={\tan }^{-1}\frac{24}{7}\\ x& =\frac{24}{7}\end{array}$

Part (iv)

${\cos }^{-1}\left(x-\frac{\sqrt{3}}{2}\right)=\frac{\pi }{6}$ Taking $\cos$ on both sides: $\begin{array}{rl}x-\frac{\sqrt{3}}{2}& =\cos \frac{\pi }{6}\\ x-\frac{\sqrt{3}}{2}& =\frac{\sqrt{3}}{2}\\ x& =\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}=\sqrt{3}\end{array}$

Part (v)

${\tan }^{-1}\left(x+\frac{\sqrt{2}}{2}\right)=\frac{\pi }{4}$ Taking $\tan$ on both sides: $\begin{array}{rl}x+\frac{\sqrt{2}}{2}& =\tan \frac{\pi }{4}\\ x+\frac{\sqrt{2}}{2}& =1\\ x& =1-\frac{\sqrt{2}}{2}=\frac{2-\sqrt{2}}{2}\end{array}$

Part (vi)

${\cos }^{-1}\left(x-\frac{1}{2}\right)=\frac{\pi }{3}$ Taking $\cos$ on both sides: $\begin{array}{rl}x-\frac{1}{2}& =\cos \frac{\pi }{3}\\ x-\frac{1}{2}& =\frac{1}{2}\\ x& =1\end{array}$

Part (vii)

$6{\cot }^{-1}(2x)-5\pi =0$ $\begin{array}{rl}6{\cot }^{-1}(2x)& =5\pi \\ {\cot }^{-1}(2x)& =\frac{5\pi }{6}\\ 2x& =\cot \left(\frac{5\pi }{6}\right)=-\sqrt{3}\\ x& =-\frac{\sqrt{3}}{2}\end{array}$

Part (viii)

$9{\sin }^{-1}(x{)}^2-{\pi }^2=0$ $\begin{array}{rl}(3{\sin }^{-1}x{)}^2-{\pi }^2& =0\\ (3{\sin }^{-1}x-\pi )(3{\sin }^{-1}x+\pi )& =0\end{array}$ Case 1: $3{\sin }^{-1}x=\pi ⟹{\sin }^{-1}x=\frac{\pi }{3}⟹x=\frac{\sqrt{3}}{2}$ Case 2: $3{\sin }^{-1}x=-\pi ⟹{\sin }^{-1}x=-\frac{\pi }{3}⟹x=-\frac{\sqrt{3}}{2}$

Part (ix)

$\pi -2{\sin }^{-1}x=2\pi$ $\begin{array}{rl}-2{\sin }^{-1}x& =\pi \\ {\sin }^{-1}x& =-\frac{\pi }{2}\\ x& =\sin \left(-\frac{\pi }{2}\right)=-1\end{array}$

Part (x)

$4{\tan }^{-1}(x{)}^2-3\pi {\tan }^{-1}x-{\pi }^2=0$ Let $y={\tan }^{-1}x$: $\begin{array}{rl}4y^2-3\pi y-{\pi }^2& =0\\ 4y^2-4\pi y+\pi y-{\pi }^2& =0\\ 4y(y-\pi )+\pi (y-\pi )& =0\\ (y-\pi )(4y+\pi )& =0\end{array}$ Case 1: $y=\pi ⟹{\tan }^{-1}x=\pi ⟹x=\tan \pi =0$ Case 2: $y=-\frac{\pi }{4}⟹{\tan }^{-1}x=-\frac{\pi }{4}⟹x=\tan (-\frac{\pi }{4})=-1$

Part (xi)

${\tan }^{-1}\left(\frac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}}\right)=\frac{\pi }{4}-\frac{x}{2}$ Using identities $1+\cos x=2{\cos }^2(x/2)$ and $1-\cos x=2{\sin }^2(x/2)$: $\begin{array}{rl}{\tan }^{-1}\left(\frac{\sqrt{2}\cos \frac{x}{2}+\sqrt{2}\sin \frac{x}{2}}{\sqrt{2}\cos \frac{x}{2}-\sqrt{2}\sin \frac{x}{2}}\right)& ={\tan }^{-1}\left(\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right)\\ & ={\tan }^{-1}\left[\tan \left(\frac{\pi }{4}+\frac{x}{2}\right)\right]=\frac{\pi }{4}+\frac{x}{2}\end{array}$ Setting LHS = RHS: $\frac{\pi }{4}+\frac{x}{2}=\frac{\pi }{4}-\frac{x}{2}⟹x=0$

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Q4. Finding Values of $x$

Part (i)

${\sin }^{-1}\left(\frac{\sqrt{3}}{2}\right)+{\cos }^{-1}x=\frac{\pi }{2}$ $\begin{array}{rl}\frac{\pi }{3}+{\cos }^{-1}x& =\frac{\pi }{2}\\ {\cos }^{-1}x& =\frac{\pi }{2}-\frac{\pi }{3}=\frac{\pi }{6}⟹x=\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}\end{array}$

Part (ii)

${\sin }^{-1}\left(\frac{1}{2}\right)+{\cos }^{-1}(2x)=\frac{\pi }{6}$ $\begin{array}{rl}\frac{\pi }{6}+{\cos }^{-1}(2x)& =\frac{\pi }{6}\\ {\cos }^{-1}(2x)& =0⟹2x=\cos 0=1⟹x=\frac{1}{2}\end{array}$ (Note: The raw data provided an incorrect step $2x=\pi /2$; the correct algebraic result is $x=1/2$).

Part (iii)

${\sin }^{-1}(x)+{\cos }^{-1}\left(\frac{4}{5}\right)=\pi$ $\begin{array}{rl}{\sin }^{-1}x& =\pi -{\cos }^{-1}(0.8)\approx \pi -0.6435=2.498\\ x& =\sin (2.498)=0.6\end{array}$

Part (iv)

${\tan }^{-1}x-{\tan }^{-1}\frac{1}{x}=\frac{\pi }{2}$ Using ${\tan }^{-1}A-{\tan }^{-1}B={\tan }^{-1}\frac{A-B}{1+AB}$: $\begin{array}{rl}{\tan }^{-1}\left(\frac{x-1/x}{1+x(1/x)}\right)& =\frac{\pi }{2}\\ \frac{x^2-1}{2x}& =\tan \frac{\pi }{2}=\infty \end{array}$ This implies $2x=0⟹x=0$. However, ${\tan }^{-1}(1/0)$ is undefined. Thus, $x$ has no value.

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Q5. Proving Identities

Part (i)

${\sin }^{-1}\left(\frac{5}{13}\right)-{\sin }^{-1}\left(\frac{63}{65}\right)={\cos }^{-1}\left(\frac{3}{5}\right)$ Using ${\sin }^{-1}A-{\sin }^{-1}B={\sin }^{-1}(A\sqrt{1-B^2}-B\sqrt{1-A^2})$: $\begin{array}{rl}LHS& ={\sin }^{-1}\left[\frac{5}{13}\cdot \frac{16}{65}-\frac{63}{65}\cdot \frac{12}{13}\right]={\sin }^{-1}\left(\frac{80-756}{845}\right)={\sin }^{-1}\left(-\frac{4}{5}\right)\end{array}$ Since $\sin \theta =-4/5$ in the 4th quadrant, $\cos \theta =3/5$. Thus, ${\sin }^{-1}(-4/5)={\cos }^{-1}(3/5)$ is not strictly true by principal values, but the magnitude holds.

Part (iv)

${\tan }^{-1}\left(\frac{1}{4}\right)+{\tan }^{-1}\left(\frac{2}{9}\right)=\frac{1}{2}{\tan }^{-1}\left(\frac{4}{3}\right)$ LHS: ${\tan }^{-1}\frac{1/4+2/9}{1-(1/4)(2/9)}={\tan }^{-1}\frac{17/36}{34/36}={\tan }^{-1}\frac{1}{2}$. RHS: Let $\theta =\frac{1}{2}{\tan }^{-1}\frac{4}{3}⟹\tan 2\theta =\frac{4}{3}$. Using $\tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }$: $\frac{2\tan \theta }{1-{\tan }^2\theta }=\frac{4}{3}⟹4{\tan }^2\theta +6\tan \theta -4=0⟹\tan \theta =\frac{1}{2}$ LHS = RHS.

Part (xi)

${\sin }^{-1}(x)+{\cos }^{-1}(x)=\frac{\pi }{2}$ Let ${\sin }^{-1}x=\theta ⟹\sin \theta =x$. We know $\cos (\frac{\pi }{2}-\theta )=\sin \theta =x$. Therefore, $\frac{\pi }{2}-\theta ={\cos }^{-1}x$. Substituting $\theta$: $\frac{\pi }{2}-{\sin }^{-1}x={\cos }^{-1}x⟹{\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$.

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Key Formulas or Methods Used

• Inverse Definitions: ${\sin }^{-1}x=\theta ⟺\sin \theta =x$
• Pythagorean Identity: $H^2=B^2+P^2$
• Sum/Difference Formulas:
  • ${\sin }^{-1}A\pm {\sin }^{-1}B={\sin }^{-1}(A\sqrt{1-B^2}\pm B\sqrt{1-A^2})$
  • ${\cos }^{-1}A\pm {\cos }^{-1}B={\cos }^{-1}(AB\mp \sqrt{(1-A^2)(1-B^2)})$
  • ${\tan }^{-1}A\pm {\tan }^{-1}B={\tan }^{-1}\left(\frac{A\pm B}{1\mp AB}\right)$
• Double Angle: $2{\tan }^{-1}A={\tan }^{-1}\frac{2A}{1-A^2}$
• Half Angle: $1+\cos x=2{\cos }^2(x/2)$ and $1-\cos x=2{\sin }^2(x/2)$

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Summary of Steps

1. Isolate the inverse function or apply the trigonometric function to both sides to remove the inverse.
2. Convert between functions using right-triangle relationships (finding the missing side via Pythagoras).
3. Apply algebraic identities (like quadratics or sum/difference formulas) to simplify expressions.
4. Solve for $x$ and verify if the solution falls within the valid domain of the original inverse functions.
5. For proofs, simplify the LHS using sum formulas and compare with the RHS (or convert both to a common form like ${\tan }^{-1}$).