Q19. If (tan⁻¹(x))² + (cot⁻¹(x))² = 5π²/8, then find x.
Answers
Step-by-step explanation:
tan
−1
x)
2
+(cot
−1
x)
2
=
8
5π
2
Since, cot^{-1}x = \frac{\pi}{2}-tan^{-1}xcot
−1
x=
2
π
−tan
−1
x
(tan^{-1}x)^2+(\frac{\pi}{2}-tan^{-1}x)^2=\frac{5\pi^2}{8}(tan
−1
x)
2
+(
2
π
−tan
−1
x)
2
=
8
5π
2
(tan^{-1}x)^2+\frac{\pi^2}{4}+(tan^{-1}x)^2-\pi tan^{-1}x=\frac{5\pi^2}{8}(tan
−1
x)
2
+
4
π
2
+(tan
−1
x)
2
−πtan
−1
x=
8
5π
2
2(tan^{-1}x)^2-\pi tan^{-1}x=\frac{5\pi^2}{8}-\frac{\pi^2}{4}2(tan
−1
x)
2
−πtan
−1
x=
8
5π
2
−
4
π
2
2(tan^{-1}x)^2-\pi tan^{-1}x=\frac{5\pi^2-2\pi^2}{8}2(tan
−1
x)
2
−πtan
−1
x=
8
5π
2
−2π
2
2(tan^{-1}x)^2-\pi tan^{-1}x=\frac{3\pi^2}{8}2(tan
−1
x)
2
−πtan
−1
x=
8
3π
2
2(tan^{-1}x)^2-\pi tan^{-1}x-\frac{3\pi^2}{8}=02(tan
−1
x)
2
−πtan
−1
x−
8
3π
2
=0
By quadratic formula,
tan^{-1}x=\frac{\pi\pm \sqrt{\pi^2+3\pi^2}}{4}tan
−1
x=
4
π±
π
2
+3π
2
tan^{-1}x=\frac{\pi\pm \sqrt{4\pi^2}}{4}tan
−1
x=
4
π±
4π
2
tan^{-1}x=\frac{\pi \pm 2\pi}{4}tan
−1
x=
4
π±2π
tan^{-1}x=\frac{3\pi}{4}tan
−1
x=
4
3π
or tan^{-1}x=-\frac{\pi}{4}tan
−1
x=−
4
π
\implies x = -1⟹x=−1
Topic :-
Inverse Trigonometric Function
Given :-
To Find :-
Value of 'x'.
Solution :-
Applying Quadratic Formula,
Put back value of p,
Answer :-