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CBSE Class 12 Math 2012 Solved Paper

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Question : 16
Total: 29
Prove that tan−1(
cosx
1+sinx
) =
π
4
−
π
2
 , x ∊ (−
π
2
,
π
2
)
OR
Prove that sin−1(
8
17
) + sin−1(
3
5
) = cos−1(
36
85
)
Solution:  
tan−1(
cosx
1+sinx
)
= tan−1 [
sin(
π
2
−x)
1+cos(
π
2
−x)
]
=
tan−1 [
2sin(
π
4
−
x
2
)cos(
π
4
−
x
2
)
2cos2(
π
4
−
x
2
)
]
[Since sin θ = 2 sin (θ/2) cos (θ/2) and 1 + cos θ = 2 cos2 (θ/2)]
tan−1 [tan(
π
4
−
x
2
)] = (
π
4
−
x
2
) (proved)
OR
Let sin−1
8
17
 = x
Then , sin x =
8
17
 ; cos x = √1−x2
⇒ cos x = √1−(
8
17
)2
⇒ cos x = √1−(
8
17
)2
⇒ cos x = √
225
289
⇒ cos x =
15
17
∴ tan x =
sinx
cosx
⇒ tan x =
8
17
15
17
⇒ tan x =
8
15
⇒ x = tan−1(
8
15
) ... (1)
Let sin−1
3
5
 = y ... (2)
Then, sin y =
3
5
 ; cos y = √1−y2
⇒ cos y = √1−(
3
5
)2
⇒ cos y = √
16
25
⇒ cos y =
4
5
∴ tan y =
siny
cosy
⇒ tan y =
3
5
4
5
⇒ tan y =
3
4
⇒ y = tan−1(
3
4
) ... (3)
From equations (2) and (3), we have,
sin−1(
3
5
) = tan−1(
3
4
)
Now consider sin−1(
8
17
) + sin−1(
3
5
)
From equations (1) and (3), we have, sin−1(
8
17
) + sin−1(
3
5
) = tan−1(
5
15
)+tan−1(
3
4
)
= tan−1(
8
15
+
3
4
1−
8
15
×
3
4
) [Since tan−1x+tan−1y = tan−1
x+y
x−y
]
= tan−1(
32+45
60−24
)
sin−1(
8
17
) + sin−1(
3
5
) = tan−1(
77
36
) ... (4)
Now, we have:
Let tan−1(
77
36
) = z
Then tan z =
77
36
⇒ sec z = √1+(
77
36
)2 [Since sec θ = √1+tan2θ]
⇒ sec z = √
1296+5929
1296
⇒ sec z = √
7225
1296
⇒ sec z =
85
36
We know that cosz =
1
secz
Thus, sec z =
85
36
 , cos z =
36
85
⇒ z = cos−1(
36
85
)
⇒ tan−1(
77
36
) = cos−1(
36
85
)
⇒ sin−1(
8
17
) + sin−1(
3
5
) = cos−1(
36
85
) [Since from equation (4)]
Hence proved.
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