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

We have (tan^-1x)² +(cot^-1x)² = $\frac { 5 \pi ^ { 2 } } { 8 }$

⇒ (tan^–1x+cot^–1x)² – 2 tan^-1x (π/2 – tan^–1x) = $\frac { 5 \pi ^ { 2 } } { 8 }$

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

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

⇒ tan^–1x = –π/4, 3π/4

⇒ tan^–1 x = –π/4 ⇒ x = –1