(secA - cosA) (secA + cosA) = sin²A + tan²A
HeY MaTeS!!!
its ReALLy confusing me very much plzz help...
Simran200211:
where i go????
what to start
achaa baad m baat krte hain
Answers
Answered by
2
Hi mate here is ur query..
************
Given LHS
=(secA - cosA) (secA + cosA)
since (a+b)(a-b)=a²-b²
hence
=sec²A-cos²A
=(1+tan²A)-(1-sin²A)
=Sin²A+tan²A=RHS
proved
***********
Hope this will help you.
************
Given LHS
=(secA - cosA) (secA + cosA)
since (a+b)(a-b)=a²-b²
hence
=sec²A-cos²A
=(1+tan²A)-(1-sin²A)
=Sin²A+tan²A=RHS
proved
***********
Hope this will help you.
thank u
pleasure☺☺
:-)
Answered by
3
(secA +cosA)(secA - cosA)
= sec²A - cos²A <difference of 2 squares>
= 1/cos²A - cos²A
= 1/cos²A - cos^4A/cos²A
= (1 - cos^4A)/cos²A
= (1-cos²A)(1+cos²A)/cos²A
= sin²A(1+cos²A)/cos²A
[sin²A/cos²A](1+cos²A)
= [sin²A/cos²A] + [sin²A/cos²A]cos²A
= sin²A/cos²A + sin²A
= tan²A + sin²A.
= sec²A - cos²A <difference of 2 squares>
= 1/cos²A - cos²A
= 1/cos²A - cos^4A/cos²A
= (1 - cos^4A)/cos²A
= (1-cos²A)(1+cos²A)/cos²A
= sin²A(1+cos²A)/cos²A
[sin²A/cos²A](1+cos²A)
= [sin²A/cos²A] + [sin²A/cos²A]cos²A
= sin²A/cos²A + sin²A
= tan²A + sin²A.
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