tan((2sin-1 (3/2) + tan-1 (5/2))
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Answer:
Let sin
−1
5
3
=x⇒sinx=
5
3
.
⇒cosx=
1−(
5
3
)
2
=
5
4
∴tanx=
4
3
∴x=tan
−1
4
3
⇒sin
−1
5
3
=tan
−1
4
3
Now, we have:
L.H.S.=2sin
−1
5
3
=2tan
−1
4
3
=tan
−1
⎝
⎜
⎜
⎜
⎜
⎛
1−(
4
3
)
2
2×
4
3
⎠
⎟
⎟
⎟
⎟
⎞
,[∵2tan
−1
x=tan
−1
1−x
2
2x
]
=tan
−1
⎝
⎜
⎜
⎛
16
16−9
2
3
⎠
⎟
⎟
⎞
=tan
−1
(
2
3
×
7
16
)
=tan
−1
7
24
=R.H.S.
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