The correct option is B 1/cosθ tan ^2θ/1+θ+1=^2θ 1 /1+θ+1 #160; #160; #160; #160; #160; #160; #160; #160; #160; #160; #160; ...

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Basic Trigonometric Identities

tan 2θ/1+θ+1 ...

Question

$\frac{{tan}^{2} θ}{1 + sec θ} + 1$ equals to

tan θ

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\frac{1}{cos θ}

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sec θ - 1

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sec θ + tan θ

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Solution

The correct option is B $\frac{1}{cos θ}$
$\frac{{tan}^{2} θ}{1 + sec θ} + 1 = \frac{{sec}^{2} θ - 1}{1 + sec θ} + 1$

$= \frac{(sec θ + 1) (sec θ - 1)}{1 + sec θ} + 1$
$= sec θ - 1 + 1$
$= sec θ$
$= \frac{1}{cos θ}$
Option B is correct.

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