Math, asked by bhaveshvk18, 1 year ago

hey

100 points solve two of them

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Answers

Answered by ravi268333
3
hi mate here is ur answer. see the 2nd option to solve the 2nd question

plz mark as brainlist answer
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bhaveshvk18: hmm..thanks bro.
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Answered by dhakatanishqddun
4
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

vIsHal005: ok ravi bro sorry
dhakatanishqddun: the talk ends here ravi is getting disturb
vIsHal005: :)
dhakatanishqddun: by the wsy im mr. not mrs.
vIsHal005: ya thats confirmed
dhakatanishqddun: never talk to some one like tha
dhakatanishqddun: that
dhakatanishqddun: rude manner
vIsHal005: what's wrong in that
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