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Byju's Answer
Standard X
Mathematics
Relation between Trigonometric Ratios
Prove that ...
Question
Prove that
sin
2
θ
cos
2
θ
+
cos
2
θ
sin
2
θ
=
sec
2
θ
−
cosec
2
θ
−
2
Open in App
Solution
L
H
S
=
s
i
n
2
θ
c
o
s
2
θ
+
c
o
s
2
θ
s
i
n
2
θ
=
s
i
n
4
θ
+
c
o
s
4
θ
s
i
n
2
θ
.
c
o
s
2
θ
We know that,
a
2
+
b
2
+
2
a
b
=
(
a
+
b
)
2
∴
(
a
+
b
)
2
−
2
a
b
=
a
2
+
b
2
=
(
s
i
n
2
θ
+
c
o
s
2
θ
)
2
−
2
s
i
n
2
θ
.
c
o
s
2
θ
s
i
n
2
θ
.
c
o
s
2
θ
=
1
−
2
s
i
n
2
θ
.
c
o
s
2
θ
s
i
n
2
θ
.
c
o
s
2
θ
=
1
s
i
n
2
θ
.
c
o
s
2
θ
−
2
=
s
e
c
2
θ
(
c
o
s
e
c
2
θ
)
−
2
(
∵
1
+
c
o
t
2
θ
=
c
o
s
e
c
2
θ
)
=
s
e
c
2
θ
(
1
+
c
o
t
2
θ
)
−
2
(
c
o
s
θ
=
1
s
e
c
θ
)
=
s
e
c
2
θ
+
s
e
c
2
θ
.
c
o
s
2
θ
s
i
n
2
θ
−
2
=
s
e
c
2
θ
+
c
o
s
e
c
2
θ
−
2
=RHS
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0
Similar questions
Q.
The minimum value of
sin
2
θ
+
cos
2
θ
+
sec
2
θ
+
c
o
s
e
c
2
θ
+
tan
2
θ
+
cot
2
θ
.
Q.
(i)
sin
6
θ
+
cos
6
θ
=
1
-
3
sin
2
θcos
2
θ
(ii)
sin
2
θ
+
cos
4
θ
=
cos
2
θ
+
sin
4
θ
(iii)
cosec
4
θ
-
cosec
2
θ
=
cot
4
θ
+
cot
2
θ
Q.
sec
2
θ
csc
2
θ
sec
2
θ
−
csc
2
θ
=
1
sin
2
θ
−
cos
2
θ
Q.
Square relations among various trigonometrical ratios
sin
2
θ
+
cos
2
θ
=
1
sec
2
θ
−
tan
2
=
1
cos
e
c
2
θ
−
cot
2
θ
=
1
These are also called Pythagorean identities.
Q.
If
tan
θ
=
√
3
−
√
2
, find
sin
2
θ
+
sec
2
θ
−
c
o
s
2
θ
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