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Inverse Trigonometric Functions
NDA 2015
Exam
NDA
Year
2015
Subject
Mathematics
Chapter
Inverse Trigonometric Functions
NDA 2015 · Mathematics · Inverse Trigonometric Functions
The equation ^ -1 (1+x)+ ^ -1 (1-x)=(π)/(2) is satisfied by
The equation
tan
−
1
(
1
+
x
)
+
tan
−
1
(
1
−
x
)
=
π
2
\tan ^{-1}(1+x)+\tan ^{-1}(1-x)=\frac{\pi}{2}
tan
−
1
(
1
+
x
)
+
tan
−
1
(
1
−
x
)
=
2
π
is satisfied by
Options
A
x
=
1
x=1
x
=
1
B
x
=
−
1
x=-1
x
=
−
1
C
x
=
0
\mathrm{x}=0
x
=
0
D
x
=
1
2
x=\frac{1}{2}
x
=
2
1
Correct answer
C.
x
=
0
\mathrm{x}=0
x
=
0
Step-by-step solution
tan
−
1
(
1
+
x
)
+
tan
−
1
(
1
−
x
)
=
π
2
tan
−
1
[
(
1
+
x
)
+
(
1
−
x
)
1
−
(
1
+
x
)
(
1
−
x
)
]
=
π
2
\begin{aligned} & \tan ^{-1}(1+x)+\tan ^{-1}(1-x)=\frac{\pi}{2} \\ & \tan ^{-1}\left[\frac{(1+x)+(1-x)}{1-(1+x)(1-x)}\right]=\frac{\pi}{2}\end{aligned}
tan
−
1
(
1
+
x
)
+
tan
−
1
(
1
−
x
)
=
2
π
tan
−
1
[
1
−
(
1
+
x
)
(
1
−
x
)
(
1
+
x
)
+
(
1
−
x
)
]
=
2
π
⇒
1
+
x
+
1
−
x
1
−
(
1
+
x
)
(
1
−
x
)
=
tan
π
2
⇒
2
1
−
(
1
+
x
)
(
1
−
x
)
=
1
0
⇒
1
−
(
1
+
x
)
(
1
−
x
)
=
0
⇒
(
1
+
x
)
(
1
−
x
)
=
1
1
−
x
2
=
1
x
2
=
0
x
=
0
\begin{array}{ll} \Rightarrow & \frac{1+x+1-x}{1-(1+x)(1-x)}=\tan \frac{\pi}{2} \\ \Rightarrow & \frac{2}{1-(1+x)(1-x)}=\frac{1}{0} \\ \Rightarrow & 1-(1+x)(1-x)=0 \\ \Rightarrow & (1+x)(1-x)=1 \\ & 1-x^2=1 \\ & x^2=0 \\ & x=0 \end{array}
⇒
⇒
⇒
⇒
1
−
(
1
+
x
)
(
1
−
x
)
1
+
x
+
1
−
x
=
tan
2
π
1
−
(
1
+
x
)
(
1
−
x
)
2
=
0
1
1
−
(
1
+
x
)
(
1
−
x
)
=
0
(
1
+
x
)
(
1
−
x
)
=
1
1
−
x
2
=
1
x
2
=
0
x
=
0
\therefore \quad$ Option (C) is correct.
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