Prove that cotA-tanA= (1-sin^2A)/sinAcosA
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Answered by
11
Hi ! .
Solution :-
From LHS
=> cotA - tanA
=> cosA / sinA - sinA / cosA
=> cos²A - sin²A / sinA * cosA
=> ( 1 - sin²A ) - sin²A / sinA * cosA
=> 1 - 2sin²A / sinA * cosA
LHS = RHS prooved ♻
______________________
Hope it helps you !!
@Rajukumar111
Solution :-
From LHS
=> cotA - tanA
=> cosA / sinA - sinA / cosA
=> cos²A - sin²A / sinA * cosA
=> ( 1 - sin²A ) - sin²A / sinA * cosA
=> 1 - 2sin²A / sinA * cosA
LHS = RHS prooved ♻
______________________
Hope it helps you !!
@Rajukumar111
Answered by
4
ᴇʟʟᴏ!!
ʜᴇʀᴇ's ʏᴏᴜʀ ᴀɴsᴡᴇʀ
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ᴛᴏ ᴘʀᴏᴠᴇ , ᴄᴏᴛ ᴀ - ᴛᴀɴ ᴀ = ( 1 - sɪɴ^2 ᴀ ) / sɪɴ ᴀ ᴄᴏs ᴀ
ɴᴏᴡ , ʟ . ʜ . s
= ᴄᴏᴛ ᴀ - ᴛᴀɴ ᴀ
= ᴄᴏs ᴀ / sɪɴ ᴀ - sɪɴ ᴀ / ᴄᴏs ᴀ
= ᴄᴏs^2 ᴀ - sɪɴ^2 ᴀ / sɪɴ ᴀ . ᴄᴏs ᴀ
= ( 1 - sɪɴ^2 ᴀ ) - sɪɴ^2 ᴀ / sɪɴ ᴀ . ᴄᴏs ᴀ
= 1 - sɪɴ^2 ᴀ - sɪɴ^2 ᴀ / sɪɴ ᴀ . ᴄᴏs ᴀ
= 1 - 2 sɪɴ^2 ᴀ / sɪɴ ᴀ . ᴄᴏs ᴀ
ᴀɴᴅ ɢɪᴠᴇɴ , ʀ . ʜ . s = ( 1 - sɪɴ^2 ᴀ ) / sɪɴᴀ . ᴄᴏsᴀ
sᴏ , ʟ . ʜ . s = ʀ . ʜ . s [ • ʜᴇɴᴄᴇ ᴘʀᴏᴠᴇᴅ ]
_______________________
ʜᴏᴘᴇ ɪᴛ ʜᴇʟᴘs ʏᴏᴜ ᴅᴇᴀʀ!
★ ᴅᴇᴠɪʟ ᴋɪɴɢ
ʜᴇʀᴇ's ʏᴏᴜʀ ᴀɴsᴡᴇʀ
______________________
ᴛᴏ ᴘʀᴏᴠᴇ , ᴄᴏᴛ ᴀ - ᴛᴀɴ ᴀ = ( 1 - sɪɴ^2 ᴀ ) / sɪɴ ᴀ ᴄᴏs ᴀ
ɴᴏᴡ , ʟ . ʜ . s
= ᴄᴏᴛ ᴀ - ᴛᴀɴ ᴀ
= ᴄᴏs ᴀ / sɪɴ ᴀ - sɪɴ ᴀ / ᴄᴏs ᴀ
= ᴄᴏs^2 ᴀ - sɪɴ^2 ᴀ / sɪɴ ᴀ . ᴄᴏs ᴀ
= ( 1 - sɪɴ^2 ᴀ ) - sɪɴ^2 ᴀ / sɪɴ ᴀ . ᴄᴏs ᴀ
= 1 - sɪɴ^2 ᴀ - sɪɴ^2 ᴀ / sɪɴ ᴀ . ᴄᴏs ᴀ
= 1 - 2 sɪɴ^2 ᴀ / sɪɴ ᴀ . ᴄᴏs ᴀ
ᴀɴᴅ ɢɪᴠᴇɴ , ʀ . ʜ . s = ( 1 - sɪɴ^2 ᴀ ) / sɪɴᴀ . ᴄᴏsᴀ
sᴏ , ʟ . ʜ . s = ʀ . ʜ . s [ • ʜᴇɴᴄᴇ ᴘʀᴏᴠᴇᴅ ]
_______________________
ʜᴏᴘᴇ ɪᴛ ʜᴇʟᴘs ʏᴏᴜ ᴅᴇᴀʀ!
★ ᴅᴇᴠɪʟ ᴋɪɴɢ
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