To prove: $\mathit{s}\mathit{i}{\mathit{n}}^{-1}\left(\frac{4}{5}\right)+\mathit{s}\mathit{i}{\mathit{n}}^{-1}\left(\frac{5}{13}\right)$ + $\mathit{s}\mathit{i}{\mathit{n}}^{-1}\left(\frac{16}{65}\right)$ = $\frac{\mathit{\pi }}{2}$
Let $\mathit{s}\mathit{i}{\mathit{n}}^{-1}\left(\frac{4}{5}\right)$ = x
⇒ sin x = $\frac{4}{5}$
⇒ cos x = $\sqrt{1-\mathit{s}\mathit{i}{\mathit{n}}^2\mathit{x}}$ = $\frac{3}{5}$
$\mathit{s}\mathit{i}{\mathit{n}}^{-1}\left(\frac{5}{13}\right)$ = y
⇒ sin y = $\frac{5}{13}$
⇒ cos y = $\sqrt{1-\mathit{s}\mathit{i}{\mathit{n}}^2\mathit{y}}$ = $\frac{12}{13}$
$\mathit{s}\mathit{i}{\mathit{n}}^{-1}\left(\frac{16}{65}\right)$ = z
⇒ sin z = $\frac{16}{65}$
⇒ cos z = $\sqrt{1-\mathit{s}\mathit{i}{\mathit{n}}^2\mathit{x}}$ = $\frac{63}{65}$
tan x = $\frac{4}{3}$ , tan y = $\frac{5}{12}$ , tan z = $\frac{16}{63}$
tan z = $\frac{16}{63}$ ⇒ cot z = $\frac{63}{16}$ ... (1)
tan (x + y) = $\frac{\mathit{t}\mathit{a}\mathit{n}\left(\mathit{x}+\mathit{y}\right)}{1-\mathit{t}\mathit{a}\mathit{n}\mathit{x}.\mathit{t}\mathit{a}\mathit{n}\mathit{y}}$
⇒ tan (x + y) = $\frac{\frac{4}{3}+\frac{5}{12}}{1-\frac{20}{36}}$
⇒ tan (x + y) = $\frac{63}{16}$
⇒ tan (x + y) = cot z ... [from equation (1)]
⇒ tan (x + y) = tan $\left(\frac{\mathit{\pi }}{2}-\mathit{z}\right)$
⇒ x + y = $\frac{\mathit{\pi }}{2}$ - z
⇒ x + y + z = $\frac{\mathit{\pi }}{2}$
∴ $\mathit{s}\mathit{i}{\mathit{n}}^{-1}\left(\frac{4}{5}\right)+\mathit{s}\mathit{i}{\mathit{n}}^{-1}\left(\frac{5}{13}\right)$ + $\mathit{s}\mathit{i}{\mathit{n}}^{-1}\left(\frac{16}{65}\right)$ = $\frac{\mathit{\pi }}{2}$
OR
$\mathit{t}\mathit{a}{\mathit{n}}^{-1}3\mathit{x}+\mathit{t}\mathit{a}{\mathit{n}}^{-1}2\mathit{x}$ = $\frac{\mathit{\pi }}{4}$
⇒ $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{5\mathit{x}}{1-6{\mathit{x}}^2}\right)$ = $\frac{\mathit{\pi }}{4}$ . 3x × 2x < 1
⇒ tan $\left[\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{5\mathit{x}}{1-6{\mathit{x}}^2}\right)\right]$ = tan $\frac{\mathit{\pi }}{4}$
⇒ $\frac{5\mathit{x}}{1-6{\mathit{x}}^2}$ = 1
⇒ 1 - $6{\mathit{x}}^2$ = 5x
⇒ $6{\mathit{x}}^2$ + 5x - 1 = 0
⇒ $6{\mathit{x}}^2$ + 6x - x - 1 = 0
⇒ x = - 1 or $\frac{1}{6}$
Here (- 3) × (- 2) ≮ 1 [Since (- 3) × (- 2) = 6 >1]
Therefore, x = 1 is not the solution.
When substituting x = $\frac{1}{6}$ in 3x 2x, we have,
3 × $\frac{1}{6}$ × 2 × $\frac{1}{6}$ = $\frac{1}{2}\times \frac{1}{3}$ = $\frac{1}{6}$ < 1
Hence x = $\frac{1}{6}$ is the solution of the given equation