Prove the following
Question:

(tanθ + 2) (2 tanθ + 1) = 5 tanθ + sec2 θ

Solution:

False

LHS $=(\tan \theta+2)(2 \tan \theta+1)$

$=2 \tan ^{2} \theta+4 \tan \theta+\tan \theta+2$

$=2\left(\sec ^{2} \theta-1\right)+5 \tan \theta+2$    $\left[\because \sec ^{2} \theta-\tan ^{2} \theta=1\right]$

$=2 \sec ^{2} \theta+5 \tan \theta=R H S$

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