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Byju's Answer
Standard XII
Mathematics
Trigonometric Ratios of Compound Angles
Prove that:i ...
Question
Prove that:
(i) tan 225° cot 405° + tan 765° cot 675° = 0
(ii)
sin
8
π
3
cos
23
π
6
+
cos
13
π
3
sin
35
π
6
=
1
2
(iii) cos 24° + cos 55° + cos 125° + cos 204° + cos 300° =
1
2
(iv) tan (−225°) cot (−405°) −tan (−765°) cot (675°) = 0
(v) cos 570° sin 510° + sin (−330°) cos (−390°) = 0
(vi)
tan
11
π
3
-
2
sin
4
π
6
-
3
4
cosec
2
π
4
+
4
cos
2
17
π
6
=
3
-
4
3
2
(vii)
3
sin
π
6
sec
π
3
-
4
sin
5
π
6
cot
π
4
=
1
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Solution
i
LHS
=
tan
225
°
cot
405
°
+
tan
765
°
cot
675
°
=
tan
90
°
×
2
+
45
°
cot
90
°
×
4
+
45
°
+
tan
90
°
×
8
+
45
°
cot
90
°
×
7
+
45
°
=
tan
45
°
cot
45
°
+
tan
45
°
-
tan
45
°
=
1
×
1
+
1
×
-
1
=
1
-
1
=
0
=
RHS
Hence
proved
.
ii
LHS
=
sin
8
π
3
cos
23
π
6
+
cos
13
π
3
sin
35
π
6
=
sin
8
3
×
180
°
cos
23
6
×
180
°
+
cos
13
3
×
180
°
sin
35
6
×
180
°
=
sin
480
°
cos
690
°
+
cos
780
°
sin
1050
°
=
sin
90
°
×
5
+
30
°
cos
90
°
×
7
+
60
°
+
cos
90
°
×
8
+
60
°
sin
90
°
×
11
+
60
°
=
cos
30
°
sin
60
°
+
cos
60
°
-
cos
60
°
=
3
2
×
3
2
+
1
2
×
-
1
2
=
3
4
-
1
4
=
2
4
=
1
2
=
RHS
Hence
proved
.
iii
LHS
=
cos
24
°
+
cos
55
°
+
cos
125
°
+
cos
204
°
+
cos
300
°
=
cos
24
°
+
cos
90
°
-
35
°
+
cos
90
°
×
1
+
35
°
+
cos
90
°
×
2
+
24
°
+
cos
90
°
×
3
+
30
°
=
cos
24
°
+
sin
35
°
-
sin
35
°
-
cos
24
°
+
sin
30
°
=
0
+
0
+
1
2
=
1
2
=
RHS
Hence
proved
.
iv
LHS
=
tan
-
225
°
cot
-
405
°
-
tan
-
765
°
cot
675
°
=
-
tan
225
°
-
cot
405
°
-
-
tan
765
°
cot
675
°
∵
tan
-
x
=
tan
x
and
cot
-
x
=
-
cot
x
=
tan
225
°
cot
405
°
+
tan
765
°
cot
675
°
=
tan
90
°
×
2
+
45
°
cot
90
°
×
4
+
45
°
+
tan
90
°
×
8
+
45
°
cot
90
°
×
7
+
45
°
=
tan
45
°
cot
45
°
+
tan
45
°
-
tan
45
°
=
1
×
1
+
1
×
-
1
=
1
-
1
=
0
=
RHS
Hence
,
proved
.
v
LHS
=
cos
570
°
sin
510
°
+
sin
-
330
°
cos
-
390
°
=
cos
570
°
sin
510
°
+
-
sin
330
°
cos
390
°
∵
sin
-
x
=
-
sin
x
and
cos
-
x
=
cos
x
=
cos
570
°
sin
510
°
-
sin
330
°
cos
390
°
=
cos
90
°
×
6
+
30
°
sin
90
°
×
5
+
60
°
-
sin
90
°
×
3
+
60
°
cos
90
°
×
4
+
30
°
=
-
cos
30
°
cos
60
°
-
-
cos
60
°
cos
30
°
=
-
cos
30
°
cos
60
°
+
cos
30
°
sin
60
°
=
0
=
RHS
Hence
proved
.
vi
LHS
=
tan
11
π
3
-
2
sin
4
π
6
-
3
4
cos
e
c
2
π
4
+
4
cos
2
17
π
6
=
tan
11
π
3
-
2
sin
4
π
6
-
3
4
cos
e
c
π
4
2
+
4
cos
17
π
6
2
=
tan
11
3
×
180
°
-
2
sin
4
6
×
180
°
-
3
4
cos
e
c
180
°
4
2
+
4
cos
17
×
180
°
6
2
=
tan
660
°
-
2
sin
120
°
-
3
4
cos
e
c
45
°
2
+
4
cos
510
°
2
=
tan
660
°
-
2
sin
120
°
-
3
4
cosec
45
°
2
+
4
cos
510
°
2
=
tan
90
°
×
7
+
30
°
-
2
sin
90
°
×
1
+
30
°
-
3
4
cosec
45
°
2
+
4
cos
90
°
×
5
+
60
°
2
=
-
cot
30
°
-
2
cos
30
°
-
3
4
cosec
45
°
2
+
4
-
sin
60
°
2
=
-
cot
30
°
-
2
cos
30
°
-
3
4
cosec
45
°
2
+
4
sin
60
°
2
=
-
3
-
2
3
2
-
3
4
2
2
+
4
3
2
2
=
-
3
-
3
-
3
2
+
3
=
3
-
4
3
2
=
RHS
Hence
proved
.
v
ii
LHS
=
3
sin
π
6
s
e
c
π
3
-
4
sin
5
π
6
c
o
t
π
4
=
3
sin
180
°
6
sec
180
°
3
-
4
sin
5
×
180
°
6
cot
180
°
4
=
3
sin
30
°
sec
60
°
-
4
sin
150
°
cot
45
°
=
3
sin
30
°
sec
60
°
-
4
sin
90
°
×
1
+
60
°
cot
45
°
=
3
sin
30
°
sec
60
°
-
4
cos
60
°
cot
45
°
=
3
×
1
2
×
2
-
4
×
1
2
×
1
=
3
-
2
=
1
=
RHS
Hence
proved
.
Suggest Corrections
0
Similar questions
Q.
Prove that:
(i) tan 720° − cos 270° − sin 150° cos 120° =
1
4
(ii) sin 780° sin 480° + cos 120° sin 150° =
1
2
(iii) sin 780° sin 120° + cos 240° sin 390 =
1
2
(iv) sin 600° cos 390° + cos 480° sin 150° = −1
(v) tan 225° cot 405° + tan 765° cot 675° = 0
Q.
Prove
tan
(
−
225
o
)
cot
(
−
405
o
)
−
tan
(
−
765
0
)
cot
(
675
o
)
=
0
Q.
Find the value of
tan
225
∘
cot
405
∘
+
tan
765
∘
cot
675
∘
Q.
The value of
cos
24
∘
+
cos
55
∘
+
cos
125
∘
+
cos
204
∘
+
cos
300
∘
is
Q.
State true or false.
cos
24
∘
+
cos
55
∘
+
cos
125
∘
+
cos
204
∘
+
cos
300
∘
=
1
2
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