AP ICET 2019 Shift 2 Paper

Question : 120 of 200

Marks: +1, -0

If tan θ =

3/4

, then tan 2θ + sec 2θ =

Solution:

\tan 2 θ+\sec2θ =\;{2 \tan θ}/{1-\tan ^{2} θ}+\;{1+\tan ^{2} θ}/{1-\tan ^{2} θ}

=\;{(1+\tan θ)^{2}}/{(1-\tan θ)(1+\tan θ)}

=\;{1+\tan θ}/{1-\tan θ}=\;{1+{3}/{4}}/{1-\;{3}/{4}}=7

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