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Byju's Answer
Standard XII
Mathematics
Integration by Substitution
Evaluate the ...
Question
Evaluate the following and justify your answer
sin
2
15
o
+
sin
2
75
o
cos
2
36
o
+
cos
2
54
o
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Solution
s
i
n
2
15
∘
+
s
i
n
2
75
∘
c
o
s
2
36
∘
+
c
o
s
2
54
∘
⇒
s
i
n
(
90
−
θ
)
=
c
o
s
θ
&
s
i
n
2
θ
+
c
o
s
2
θ
=
1
&
c
o
s
(
90
−
θ
)
=
s
i
n
θ
⇒
s
i
n
2
(
90
−
75
)
+
s
i
n
2
75
∘
c
o
s
2
(
90
−
54
)
+
c
o
s
2
54
⇒
c
o
s
2
75
+
s
i
n
2
75
s
i
n
2
54
+
c
o
s
2
54
=
1
1
=
1
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