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Byju's Answer
Standard XII
Mathematics
Trigonometric Ratios of Compound Angles
Prove that:i ...
Question
Prove that:
(i)
sinθ
cos
(
90
°
-
θ
)
+
sin
(
90
°
-
θ
)
cosθ
=
1
(ii)
sinθ
cos
(
90
°
-
θ
)
+
cosθ
sin
(
90
°
-
θ
)
=
2
(iii)
sinθ
cos
(
90
°
-
θ
)
cosθ
sin
(
90
°
-
θ
)
+
cosθ
sin
(
90
°
-
θ
)
sinθ
cos
(
90
°
-
θ
)
=
1
(iv)
cos
(
90
°
-
θ
)
sec
(
90
°
-
θ
)
tanθ
cosec
(
90
°
-
θ
)
sin
(
90
°
-
θ
)
cot
(
90
°
-
θ
)
+
tan
(
90
°
-
θ
)
cotθ
=
2
(v)
cos
(
90
°
-
θ
)
1
+
sin
(
90
°
-
θ
)
+
1
+
sin
(
90
°
-
θ
)
cos
(
90
°
-
θ
)
=
2
cosecθ
(vi)
sec
90
°
-
θ
cosecθ
-
tan
90
°
-
θ
cotθ
+
cos
2
25
°
+
cos
2
65
°
3
tan
27
°
tan
63
°
=
2
3
CBSE
2010
(vii)
cotθ
tan
90
°
-
θ
-
sec
90
°
-
θ
cosecθ
+
3
tan
12
°
tan
60
°
tan
78
°
=
2
CBSE
2010
Open in App
Solution
(
i
)
LH
S
=
sin
θ
cos
(
90
0
−
θ
)
+
sin
(
90
0
−
θ
)
cos
θ
=
sinθsinθ
+
cosθcosθ
=
sin
2
θ
+
cos
2
θ
=
1
=
RHS
Hence
proved
.
(
ii
)
LHS
=
sinθ
cos
(
90
0
−
θ
)
+
cosθ
sin
(
90
0
−
θ
)
=
sin
θ
sin
θ
+
cos
θ
cos
θ
=
1
+
1
=
2
=
RHS
Hence
pr
o
v
e
d
.
(
iii
)
LHS
=
sin
θ
cos
(
90
0
−
θ
)
cos
θ
sin
(
90
0
−
θ
)
+
cos
θ
sin
(
90
0
−
θ
)
sin
θ
cos
(
90
0
−
θ
)
=
sin
θ
sin
θ
cos
θ
cos
θ
+
cos
θ
cos
θ
sin
θ
sin
θ
=
sin
2
θ
+
cos
2
θ
=
1
=
RHS
Hence
proved
(
iv
)
LHS
=
cos
(
90
0
−
θ
)
sec
(
90
0
−
θ
)
tan
θ
cos
ec
(
90
0
−
θ
)
sin
(
90
0
−
θ
)
cot
(
90
0
−
θ
)
+
tan
(
90
0
−
θ
)
cot
θ
=
sin
θ
cos
ec
θ
tan
θ
sec
θ
cos
θ
tan
θ
+
cot
θ
cot
θ
=
1
+
1
=
2
=
RHS
H
e
n
c
e
pr
o
v
e
d
.
(
v
)
LHS
=
cos
(
90
0
−
θ
)
1
+
sin
(
90
0
−
θ
)
+
1
+
sin
(
90
0
−
θ
)
cos
(
90
0
−
θ
)
=
sin
θ
1
+
cos
θ
+
1
+
cos
θ
sin
θ
=
sin
2
θ
+
(
1
+
cos
θ
)
2
(
1
+
cos
θ
)
sin
θ
=
sin
2
θ
+
1
+
cos
2
θ
+
2
cos
θ
(
1
+
cos
θ
)
sin
θ
=
1
+
1
+
2
cos
θ
(
1
+
cos
θ
)
sin
θ
=
2
+
2
cos
θ
(
1
+
cos
θ
)
sin
θ
=
2
(
1
+
cos
θ
)
(
1
+
cos
θ
)
sin
θ
=
2
1
sin
θ
=
2
cos
ec
θ
=
RHS
H
e
n
c
e
pr
o
v
e
d
.
vi
LHS
=
sec
90
°
-
θ
cosec
θ
-
tan
90
°
-
θ
cot
θ
+
cos
2
25
°
+
cos
2
65
°
3
tan
27
°
tan
63
°
=
cosec
θ
cosec
θ
-
cot
θ
cot
θ
+
sin
2
90
°
-
25
°
+
cos
2
65
°
3
tan
27
°
cot
90
°
-
63
°
=
cosec
2
θ
-
cot
2
θ
+
sin
2
65
°
+
cos
2
65
°
3
tan
27
°
cot
27
°
=
1
+
1
3
×
tan
27
°
×
1
tan
27
°
=
2
3
=
RHS
vii
LHS
=
cot
θ
tan
90
°
-
θ
-
sec
90
°
-
θ
cosec
θ
+
3
tan
12
°
tan
60
°
tan
78
°
=
cot
θ
cot
θ
-
cosec
θ
cosec
θ
+
3
tan
12
°
×
3
×
cot
90
°
-
78
°
=
cot
2
θ
-
cosec
2
θ
+
3
tan
12
°
cot
12
°
=
-
1
+
3
×
tan
12
°
×
1
tan
12
°
=
-
1
+
3
=
2
=
RHS
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