11. What is the value of sin⁴O-cos⁴O
sin²O-cos²O
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Answered by
5
Answer:
We, have
LHS = sin
4
A + cos
4
A
⇒ LHS = (sin
2
A)
2
+(cos
2
A)
2
+2sin
2
Acos
2
A−2sin
2
Acos
2
A [Adding and subtracting 2 sin
2
A cos
2
A ]
⇒ LHS = (sin
2
A+cos
2
A)
2
−2sin
2
Acos
2
A=1−2sin
2
Acos
2
A=RHS
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