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Byju's Answer
Standard X
Mathematics
Range of Trigonometric Ratios from 0 to 90 Degrees
Prove each of...
Question
Prove each of the following identities:
i
1
+
sin
θ
1
-
sin
θ
=
sec
θ
+
tan
θ
ii
1
-
cos
θ
1
+
cos
θ
=
cosec
θ
-
cot
θ
iii
1
+
cos
θ
1
-
cos
θ
+
1
-
cos
θ
1
+
cos
θ
=
2
cosec
θ
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Solution
i
LHS
=
1
+
sin
θ
1
-
sin
θ
=
1
+
sin
θ
1
-
sin
θ
×
1
+
sin
θ
1
+
sin
θ
=
1
+
sin
θ
2
1
-
sin
2
θ
=
1
+
sin
θ
2
cos
2
θ
=
1
+
sin
θ
cos
θ
=
1
cos
θ
+
sin
θ
cos
θ
=
sec
θ
+
tan
θ
=
RHS
ii
LHS
=
1
-
cos
θ
1
+
cos
θ
=
1
-
cos
θ
1
+
cos
θ
×
1
-
cos
θ
1
-
cos
θ
=
1
-
cos
θ
2
1
-
cos
2
θ
=
1
-
cos
θ
2
sin
2
θ
=
1
-
cos
θ
sin
θ
=
1
sin
θ
-
cos
θ
sin
θ
=
cosec
θ
-
cot
θ
=
RHS
(
iii
)
LHS=
1
+
cos
θ
1
−
cos
θ
+
1
−
cos
θ
1
+
cos
θ
=
(
1
+
cos
θ
)
2
(
1
−
cos
θ
)
(
1
+
cos
θ
)
+
(
1
−
cos
θ
)
2
(
1
+
cos
θ
)
(
1
−
cos
θ
)
=
(
1
+
cos
θ
)
2
(
1
−
cos
2
θ
)
+
(
1
−
cos
θ
)
2
(
1
−
cos
2
θ
)
=
(
1
+
cos
θ
)
2
sin
2
θ
+
(
1
−
cos
θ
)
2
sin
2
θ
=
(
1
+
cos
θ
)
sin
θ
+
(
1
−
cos
θ
)
sin
θ
=
1
+
cos
θ
+1
−
cos
θ
sin
θ
=
2
sin
θ
=2cosec
θ
=RHS
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0
Similar questions
Q.
(i)
1
-
sinθ
1
+
sinθ
=
(
secθ
-
tanθ
)
2
(ii)
1
+
cosθ
1
-
cosθ
=
(
cosecθ
+
cotθ
)
2
Q.
Prove the following trigonometric identities.
(i)
sec
θ
-
1
sec
θ
+
1
+
sec
θ
+
1
sec
θ
-
1
=
2
cosec
θ
(ii)
1
+
sin
θ
1
-
sin
θ
+
1
-
sin
θ
1
+
sin
θ
=
2
sec
θ
(iii)
1
+
cos
θ
1
-
cos
θ
+
1
-
cos
θ
1
+
cos
θ
=
2
cosec
θ
(iv)
sec
θ
-
1
sec
θ
+
1
=
sin
θ
1
+
cos
θ
2
(v)
sin
θ
+
1
-
cos
θ
cos
θ
-
1
+
sin
θ
=
1
+
sin
θ
cos
θ