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CBSE Class 12 Math 2013 Solved Paper

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Question : 12
Total: 29
Find the value of the following:
tan
1
2
|sin−1
2x
1+x2
+cos−1
1−y2
1+y2
| , |x| < 1, y > 0 and xy < 1
OR
Prove that tan−1(
1
2
) + tan−1(
1
5
) + tan−1(
1
8
) =
π
4
Solution:  
We know that:
sin−1
2x
1+x2
 = 2 tan−1 x for |x| ≤ 1 .. (1)
cos−1
1−y2
1+y2
 = 2 tan−1 y for y = 0 ... (2)
∴ sin−1
2x
1+x2
 + cos−1
1−y2
1+y2
 = 2tan−1 x + 2 tan−1 y
⇒ tan
1
2
|sin−1
2x
1+x2
+cos−1
1−y2
1+y2
|
= tan
1
2
 (2 tan−1 x + 2 tan−1 y)
= tan (tan−1x+tan−1y)
= tan (tan−1
x+y
1−xy
)
[Since tan−1x+tan−1y = tan−1
x+y
1−xy
 , for xy < 1]
=
x+y
1−xy
OR
We know that:
tan−1x+tan−1y = tan−1
x+y
1−xy
 , for xy < 1
We have:
tan−1(
1
2
) + tan−1(
1
5
) + tan−1(
1
8
)
= |tan−1(
1
2
) + tan−1(
1
5
)| + tan−1(
1
8
)
= tan−1(
1
2
+
1
5
1−
1
2
×
1
5
) + tan−1(
1
8
) (Since
1
2
×
1
5
 <1)
= tan−1(
7
9
)+tan−1(
1
8
)
= tan−1
7
9
+
1
8
1−
7
9
×
1
8
= tan−1
56+9
72−7
 (Since
7
9
×
1
8
 < 1)
= tan−1
65
65
 = tan−1 1 =
π
4
Hence, tan−1(
1
2
) + tan−1(
1
5
) + tan−1(
1
8
) =
π
4
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