ans this ques if it genius
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7
hello.. ☺
sᴇᴄ4ᴀ - sᴇᴄ2ᴀ
«« take theta as A »»
= sᴇᴄ2ᴀ (sᴇᴄ2ᴀ-1)
[ᴛᴀᴋɪɴɢ sᴇᴄ2ᴀ ᴄᴏᴍᴍᴏɴ]
= (1+ᴛᴀɴ2ᴀ) [(1+ᴛᴀɴ2ᴀ) - 1]
[sᴇᴄ2ᴀ-ᴛᴀɴ2ᴀ=1 sᴇᴄ2ᴀ=1+ᴛᴀɴ2ᴀ]
= (1+ᴛᴀɴ2ᴀ) [1 + ᴛᴀɴ2ᴀ - 1]
= (1+ᴛᴀɴ2ᴀ) [1 + ᴛᴀɴ2ᴀ - 1]
= (1+ᴛᴀɴ2ᴀ)ᴛᴀɴ2ᴀ
= ᴛᴀɴ2ᴀ + ᴛᴀɴ4ᴀ
= ᴛᴀɴ4ᴀ + ᴛᴀɴ2ᴀ
thank you ♥
sᴇᴄ4ᴀ - sᴇᴄ2ᴀ
«« take theta as A »»
= sᴇᴄ2ᴀ (sᴇᴄ2ᴀ-1)
[ᴛᴀᴋɪɴɢ sᴇᴄ2ᴀ ᴄᴏᴍᴍᴏɴ]
= (1+ᴛᴀɴ2ᴀ) [(1+ᴛᴀɴ2ᴀ) - 1]
[sᴇᴄ2ᴀ-ᴛᴀɴ2ᴀ=1 sᴇᴄ2ᴀ=1+ᴛᴀɴ2ᴀ]
= (1+ᴛᴀɴ2ᴀ) [1 + ᴛᴀɴ2ᴀ - 1]
= (1+ᴛᴀɴ2ᴀ) [1 + ᴛᴀɴ2ᴀ - 1]
= (1+ᴛᴀɴ2ᴀ)ᴛᴀɴ2ᴀ
= ᴛᴀɴ2ᴀ + ᴛᴀɴ4ᴀ
= ᴛᴀɴ4ᴀ + ᴛᴀɴ2ᴀ
thank you ♥
NidhraNair:
sat
say
a: tan^2ø Sec^ø
b: tan^2ø/SEC^2 Ø
c:cosec²∅ cot²∅
d:cot²∅/cosec²∅
answer is A
pls answer sisyy
Ohk I'll try :)
answer plz
Answered by
5
sec^4¢-sec^2¢
(1+tan^2¢)^2 -(1+tan^2¢)
take common (1+tan^2¢)
(1+tan^2¢)[1+tan^2¢ -1]
(1+tan^2¢)(tan^2¢)
=sec^2¢tan^2¢
hope i am right .............
(1+tan^2¢)^2 -(1+tan^2¢)
take common (1+tan^2¢)
(1+tan^2¢)[1+tan^2¢ -1]
(1+tan^2¢)(tan^2¢)
=sec^2¢tan^2¢
hope i am right .............
ur correct brw
thnx
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