Its Urgent...Help Me
Attachments:
Answers
Answered by
3
LHS = (CotA -1) /( 2- Sec²A )
= ( CotA-1) / (1 + 1 - Sec²A)
= (1/tanA)-1 / ( 1 - tan²A) ●Since tan²A +1 = sec²A So, 1- Sec²A = - tan²A
= (1-tanA) / tanA (1+tanA)(1-tanA) ● by identity a² -b² = (a+b) ( a-b)
= 1/ tanA(1+tanA)
= CotA / (1+tanA) = RHS
[ Hence Proved]
Hope this helps you friend
Thanks ✌️ ✌️
= ( CotA-1) / (1 + 1 - Sec²A)
= (1/tanA)-1 / ( 1 - tan²A) ●Since tan²A +1 = sec²A So, 1- Sec²A = - tan²A
= (1-tanA) / tanA (1+tanA)(1-tanA) ● by identity a² -b² = (a+b) ( a-b)
= 1/ tanA(1+tanA)
= CotA / (1+tanA) = RHS
[ Hence Proved]
Hope this helps you friend
Thanks ✌️ ✌️
pavamuruganpbcfes:
that's why
oooooo
Hmm
wait for 5 min
Why??
kch kaam tha?
or btao apne bare me kch
Talk in English please
??? ok bye
kkkkk bye
Answered by
4
Step-by-step explanation:
To prove that [(cotA-1)/(2-sec^2A)] = cotA/(1+tanA)
LHS = [(cotA-1)/(2-sec^2A)]
= [(cotA-1)/(2-{1 + tan^2A}]
= [(cotA-1)/(1 - tan^2A)
= [(cotA-1)/(1+tanA)(1-tanA)
=cotA [1–1/cotA)/(1+tanA)(1-tanA)
=cotA [1–tanA)/(1+tanA)(1-tanA)
=cotA/(1+tanA). Proved
hope its help you
thanx and be brainly........................................
plz mark as brainlist answer and follow me
Similar questions