prove it
Sin²A - cos²B = sin²B - cos²A
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Answer:
in
2
Acos
2
B−cos
2
Asin
2
B=sin
2
A−sin
2
B , proved.
Step-by-step explanation:
To prove that, \sin^2A \cos^2B-\cos^2A \sin^2 B = \sin^2A-\sin^2Bsin
2
Acos
2
B−cos
2
Asin
2
B=sin
2
A−sin
2
B .
L.H.S. = \sin^2A \cos^2B-\cos^2A \sin^2 Bsin
2
Acos
2
B−cos
2
Asin
2
B
Using the trigonometric identity,
\sin^2\theta+ \cos^2\thetasin
2
θ+cos
2
θ = 1
⇒ \cos^2\thetacos
2
θ = 1 - \sin^2\thetasin
2
θ
= \sin^2A (1-\sin^2 B)-(1-\sin^2 A) \sin^2 Bsin
2
A(1−sin
2
B)−(1−sin
2
A)sin
2
B
= \sin^2A -\sin^2A\sin^2 B-\sin^2 B+\sin^2 A \sin^2 Bsin
2
A−sin
2
Asin
2
B−sin
2
B+sin
2
Asin
2
B
= \sin^2A-\sin^2Bsin
2
A−sin
2
B
= L.H.S., proved.
Thus, \sin^2A \cos^2B-\cos^2A \sin^2 B = \sin^2A-\sin^2Bsin
2
Acos
2
B−cos
2
Asin
2
B=sin
2
A−sin
2
B , proved.
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