Prove that cot -1 pq +1 / p - q + - Maths - Inverse Trigonometric Functions

Question

Prove that cot -1 pq +1 / p - q + cot -1
qr +1 / q - r + cot -1 rp + 1 / r - p = 0

Shridhar Kaushik · Accepted Answer

We know that â€“ $\tan^{- 1} (\frac{1}{x}) = {\begin{matrix} \cot^{- 1} x; for x > 0 \\ - π + \cot^{- 1} x; for x < 0 \end{matrix} \Rightarrow \cot^{- 1} x = {\begin{matrix} \tan^{- 1} \frac{1}{x}; for x > 0 \\ π + \tan^{- 1} \frac{1}{x}; for x < 0 \end{matrix} ∴ \cot^{- 1} (\frac{p q + 1}{p - q}) + \cot^{- 1} (\frac{q r + 1}{q - r}) + \cot^{- 1} (\frac{r p + 1}{r - p}) [\begin{matrix} ∴ r < p \\ ∴ r - p is negative \end{matrix}] = \tan^{- 1} (\frac{p - q}{p q + 1}) + \tan^{- 1} (\frac{q - r}{q r + 1}) + π + \tan^{- 1} (\frac{r - p}{r p + 1}) = (\tan^{- 1} p - \tan^{- 1} q) + (\tan^{- 1} q - \tan^{- 1} r) + π + (\tan^{- 1} r - \tan^{- 1} p) = π$ So, the R H S should be $π$ in the query at the place of 0 and the additional information should be added that $p > q > r > 0$

Prove that cot -1 pq +1 / p - q + cot -1 qr +1 / q - r + cot -1 rp + 1 / r - p = 0

Prove that cot ^-1 pq +1 / p - q + cot ^-1 qr +1 / q - r + cot ^-1 rp + 1 / r - p = 0