sin inverse 4/5 +2 tan inverse 1/3
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2
Answer:
Toprovesin
−1
(
5
4
)+2tan
−1
(
3
1
)=
2
π
now,
L.H.S=sin
−1
(
5
4
)+2tan
−1
(
3
1
)
=sin
−1
(
5
4
)+tan
−1
[
1−(
3
1
)
2
2×
3
1
][∴2tan
−1
x=
1−x
2
2x
]
=sin
−1
(
5
4
)+tan
−1
(
8/9
2/3
)
=sin
−1
(
5
4
)+tan
−1
(
4
3
)
=tan
−1
[
1−(
5
4
)
2
5
4
]+tan
−1
(
4
3
)[∴sin
−1
x=tan
−1
(
1−x
2
x
)]
=tan
−1
(
5
4
)+tan
−1
(
5
4
)
=tan
−1
(
5
4
)+cot
−1
(
3
4
)[as,tan
−1
(
x
1
)=cot
−1
x]
=
2
π
[∴tan
−
x+cot
−1
x=
2
π
,forallx∈R]
=R.H.S
Hence,
L.H.S=R.H.Sproved
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