2sin^2A - cos^A = 2
Find the value of A
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Answered by
0
2sin^2A-cos^2A=2
1-cosA-(1+cosA÷2)=2
2-2cosA-1-cosA=4
1-3cosA=4
-3cosA=3
-cosA=1
cosA=-1
CosA=cosπ
A=π
1-cosA-(1+cosA÷2)=2
2-2cosA-1-cosA=4
1-3cosA=4
-3cosA=3
-cosA=1
cosA=-1
CosA=cosπ
A=π
Answered by
3
⇒ 2 sin²A - cos²A = 2
⇒ 2 sin²A - 2 = cos²A
⇒ 2(sin²A - 1) = cos²A
As sin²A - 1 = -cos²A from identity sin²A + cos²A = 1
⇒ 2(-cos²A) = cos²A
⇒ -2 cos²A = cos²A
⇒ 3 cos²A = 0
⇒ cos²A = 0/3 ⇒ cos²A = 0
As from trigonometric table cos 90° = 0 ⇒ cos²90° = 0
cos²A = cos²90°
Cancelling cos²A on both sides
A = 90°
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☺ ☺ ☺ Hope this Helps ☺ ☺ ☺
⇒ 2 sin²A - 2 = cos²A
⇒ 2(sin²A - 1) = cos²A
As sin²A - 1 = -cos²A from identity sin²A + cos²A = 1
⇒ 2(-cos²A) = cos²A
⇒ -2 cos²A = cos²A
⇒ 3 cos²A = 0
⇒ cos²A = 0/3 ⇒ cos²A = 0
As from trigonometric table cos 90° = 0 ⇒ cos²90° = 0
cos²A = cos²90°
Cancelling cos²A on both sides
A = 90°
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☺ ☺ ☺ Hope this Helps ☺ ☺ ☺
nitthesh7:
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Thanks for answering my question and Welcome.
No mention I can even help u with some other queries
Can you, please, answer the latest question asked by shaili123456? The question is of trigonometry.
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