Question
(sinA+cosecA)2−(sinA−cosecA)2=
Answer: Option D
:
D
⇒(sinA+cosec A)2−(sinA−cosec A)2
=4sinAcosec A
as[(a+b)2−(a−b)2=4ab]
now, (sinA×cosec A=sinA×1sinA=1)
∴4.sinAcosec A=4×1=4
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:
D
⇒(sinA+cosec A)2−(sinA−cosec A)2
=4sinAcosec A
as[(a+b)2−(a−b)2=4ab]
now, (sinA×cosec A=sinA×1sinA=1)
∴4.sinAcosec A=4×1=4
Was this answer helpful ?
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