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Question

(1+tanαtanβ)2+(tanαtanβ)2=sec2αsec2β


Solution

LHS=(1+tan α tan β)2+(tan αtan β)2

=1+(tanα+tanβ)2+2.1tanαtanβ+(tanα)2+(tanβ)2+2tanα.tanβ

[Using (a+b)2=a2+b2+2aband(ab)2=a2+b22ab]

=1+tan2αtan2β+2tanαtanβ+tan2α+tan2β2tanαtanβ

=1+tan2α+tan2αtan2β+tan2β

=sec2α+tan2β(1+tan2α)[1+tan2α=sec2α]

=sec2α+tan2β.sec2α

=sec2α(1+tan2β)

=sec2α.sec2β[1+tan2β=sec2β]

=RHS Hence proved.

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