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Question

(secAsecB+tanAtanB)2(secAtanB+tanAsecB)2=1

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Solution

LHS=(secAsecB+tanAtanB)2(secAtanB+tanAsecB)2

=(secAsecB)2+(tanAtanB)2 +2 sec A sec B tan A tan B

=[(secAtanB)2(tanAsecB)2+sec A tan B tan A sec B]

[Using (a+b)2=a2+b2+2ab]

=sec2Asec2B+tan2Atan2 B +2sec A sec B tan A tan B

sec2Atan2Btan2Asec2 B -2sec A sec B tan A tan B [Using (ab)2=a2b2]

=sec2Asec2Bsec2Atan2B+tan2Atan2Btan2Asec2B.

=sec2A(sec2Btan2B)+tan2A(tan2Bsec2B)

=sec2A1tan2A1[sec2θ=1+tan2θsec2θtan2θ=1]

=1+tan2Atan2A=1

=RHS Hence proved.


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