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UPSC NDA Math Model Paper 8

Question : 29 of 120

Marks: +1, -0

What is

(1+\tan\alpha\tan\beta)^{2}+(\tan\alpha-\tan\beta)^{2}

-\sec^{2}\alpha\sec^{2}\beta

equal to?

Solution:

\Rightarrow(1+\tan\alpha\tan\beta)^{2}=1+\tan^{2}\alpha\tan^{2}\beta

+2\tan\alpha\tan\beta

........(1)

\Rightarrow(\tan\alpha-\tan\beta)^{2}=\tan^{2}\alpha+\tan^{2}\beta

-2\tan\alpha\tan\beta

........(2)

Adding (1) and (2), we get

\Rightarrow(1+\tan\alpha\tan\beta)^{2}+(\tan\alpha-\tan\beta)^{2}

=1+\tan^{2}\alpha\tan^{2}\beta+2\tan\alpha\tan\beta+\tan^{2}\alpha+\tan^{2}\beta

-2\tan\alpha\tan\beta

=1+\tan^{2}\alpha\tan^{2}\beta+\tan^{2}\alpha+\tan^{2}\beta

=1+\tan^{2}\alpha+\tan^{2}\alpha\tan^{2}\beta+\tan^{2}\beta

=(1+\tan^{2}\alpha)+\tan^{2}\beta(1+\tan^{2}\alpha)

=(1+\tan^{2}\alpha)(1+\tan^{2}\beta)

=\sec^{2}\alpha\sec^{2}\beta

\Rightarrow(1+\tan\alpha\tan\beta)^{2}+(\tan\alpha-\tan\beta)^{2}

-\sec^{2}\alpha\sec^{2}\beta=0

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