Inverse Trigonometric Functions Worksheets With Answers Page 7
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MATHEMATICS
Long Answer (L.A.)
π
3
17
Prove that 2sin
–1
– tan
–1
=
Example 13
5
31
4
−π π
⎡
⎤
3
3
,
= θ, then sinθ =
, where θ ∈
Let sin
–1
⎢
⎥
Solution
⎣
2
2
⎦
5
5
3
3
Thus tan θ =
, which gives θ = tan
.
–1
4
4
3
17
Therefore,
2sin
– tan
–1
–1
5
31
17
3
17
= 2θ – tan
= 2 tan
– tan
–1
–1
–1
31
4
31
⎛
⎞
3
2.
⎜
⎟
17
4
–1
–1
24
17
tan
– tan
⎜
⎟
−
–1
tan
=
9
= tan
–1
31
⎜
⎟
1–
7
31
⎝
⎠
16
⎛
⎞
24 17
−
⎜
⎟
7
31
π
–1
tan
⎜
⎟
=
=
24 17
⎜
⎟
+
4
1
.
⎝
⎠
7 31
Prove that
Example 14
cot
7 + cot
8 + cot
18 = cot
3
–1
–1
–1
–1
We have
Solution
cot
7 + cot
8 + cot
18
–1
–1
–1
1
1
1
1
= tan
+ tan
+ tan
(since cot
x = tan
, if x > 0)
–1
–1
–1
–1
–1
7
8
18
x
⎛
1 1
⎞
+
⎜
⎟
1
7 8
1 1
–1
–1
+
tan
tan
⎜
⎟
.
=
(since x . y =
< 1)
1 1
18
⎜
⎟
− ×
7 8
1
⎝
⎠
7 8
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