hey
100 points solve two of them
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hi mate here is ur answer. see the 2nd option to solve the 2nd question
plz mark as brainlist answer
plz mark as brainlist answer
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dhakatanishqddun:
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hey bro ....
its 2 que.' solution
LHS=1+secAsecA=1secA+secAsecA
=cosA+1
[∵1secA=cosA]
=1+cosA
Multiply and divide by the conjugate of the term, here, by 1−cosA
=(1+cosA)×(1−cosA)(1−cosA)
=(1+cosA)(1−cosA)1−cosA
=1−cos2A1−cosA
[∵(a+b)(a−b)=a2−b2]
=sin2A1−cosA=RHS
[∵1−cos2A=sin2A].
LHS from RHS is even easier.
RHS=sin2A1−cosA
=1−cos2A1−cosA
[∵sin2A=1−cos2A].
=(1−cosA)(1+cosA)1−cosA
[∵(a−b)(a+b)=a2−b2]
=1+cosA
[∵By cancelling 1−cosA]
=1+1secA
[∵cosA=1secA]
mark it as brainlist plz bro
its 2 que.' solution
LHS=1+secAsecA=1secA+secAsecA
=cosA+1
[∵1secA=cosA]
=1+cosA
Multiply and divide by the conjugate of the term, here, by 1−cosA
=(1+cosA)×(1−cosA)(1−cosA)
=(1+cosA)(1−cosA)1−cosA
=1−cos2A1−cosA
[∵(a+b)(a−b)=a2−b2]
=sin2A1−cosA=RHS
[∵1−cos2A=sin2A].
LHS from RHS is even easier.
RHS=sin2A1−cosA
=1−cos2A1−cosA
[∵sin2A=1−cos2A].
=(1−cosA)(1+cosA)1−cosA
[∵(a−b)(a+b)=a2−b2]
=1+cosA
[∵By cancelling 1−cosA]
=1+1secA
[∵cosA=1secA]
mark it as brainlist plz bro
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