1+cosx /1- cosx= ( cosecx + cotx) ^2
Answers
Answered by
0
Answer:
Explanation:
(
csc
−
cot
x
)
2
=
1
−
cos
x
1
+
cos
x
(
csc
2
x
−
2
cot
x
csc
x
+
cot
2
x
)
=
1
−
cos
x
1
+
cos
x
(
1
sin
2
x
−
2
cos
x
sin
2
x
+
cos
2
x
sin
2
x
)
=
1
−
cos
x
1
+
cos
x
(
1
−
2
cos
x
+
cos
2
x
sin
2
x
)
=
1
−
cos
x
1
+
cos
x
(
(
1
−
cos
x
)
(
1
−
cos
x
)
1
−
cos
2
x
)
=
1
−
cos
x
1
+
cos
x
(
1
−
cos
x
(
1
−
cos
x
)
1
−
cos
x
(
1
+
cos
x
)
)
=
1
−
cos
x
1
+
cos
x
1
−
cos
x
1
+
cos
x
=
1
−
cos
x
1
+
cos
x
Answered by
0
Answer:
L. H. S
1+cos x/1- cos x
=(1+cos x/1-cos x) ×(1+cos x/1+ cos x)
(by rationalising)
= (1+cos x) ^2/1-cos^2x
(using algebraic identities)
=1+cos^2x+2cos x/sin^2x
( sin^2 x+cos^2x=1)
=( 1/ sin^2x) +(cos^2x/sin^2x)+(2 cos X x / sin x sinx)
( by dividing all terms by sin^2x individually & We can write sin^2x as sin x sin x)
= cosec^2x + cot ^2x + 2 cot x cosec x
( cosec x= 1/ sin x, cos x / sin x= cot x)
= ( cosec x+ cot x) ^2
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