19 cot
1- Sino 1- Sina
1+ sinx-1- sinx
ve (0.5
2
im
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1
Answer:
We have,
⇒cot
−1
(
1−sinx
−
1+sinx
1−sinx
+
1+sinx
)
⇒cot
−1
(
(1−sinx)−(1+sinx)
(
1−sinx
+
1+sinx
)
2
)
⇒cot
−1
(
1−sinx−1−sinx
1−sinx+1+sinx+2
1−sin
2
x
)
⇒cot
−1
(
−2sinx
2+2
cos
2
x
)
⇒cot
−1
(
−sinx
1+cosx
)
⇒cot
−1
⎝
⎛
−2sin
2
x
cos
2
x
1+2cos
2
x/2−1
⎠
⎞
⇒cot
−1
(
sinx/2
cosx/2
)
⇒cot
−1
(−cotx/2)
⇒cot
−1
(cot(π−x/2))
⇒π−x/2.
Step-by-step explanation:
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