Question

If (tan^–1x)² + (cot^–1x)² = $\frac { 5 \pi ^ { 2 } } { 8 }$ , then x equals-

• A. – 1
• B. 1
• C. 0
• D. None of these

Answer 1

Answer: (A)

– 1

Answer 2

**Sol. (**tan^–1x + cot^–1x) **–**2tan^–1x $\left( \frac { \pi } { 2 } - \tan ^ { - 1 } x \right)$ = $\frac { 5 \pi ^ { 2 } } { 8 }$

= $\frac { \pi ^ { 2 } } { 4 }$ – 2 $\left( \frac { \pi } { 2 } \right)$ tan^–1x + 2(tan^–1x)² = $\frac { 5 \pi ^ { 2 } } { 8 }$

2(tan^–1x)² – p(tan^–1x) – $\frac { 3 \pi ^ { 2 } } { 8 }$ = 0

tan^–1x = – $\frac { \pi } { 4 }$, $\frac { 3 \pi } { 4 }$Ž tan^–1x = – $\frac { \pi } { 4 }$ ŽQ x = – 1