Which of the following options is equal to sin θ + 1 − cos θ cos θ − 1 + sin θ ? Options: A . &nbsp cos θ 1 − sin θ B . &nbsp 1 + sin θ cos θ C . &nbsp 1 + sin θ 1 − sin θ D . &nbsp tan θ + sec θ

Answer: Option A : A, B, and D s i n θ + 1 − c o s θ c o s θ − 1 + s i n θ Dividing the numerator and denominator by c o s θ we get, s i n θ c o s θ + 1 c o s θ − c o s θ c o s θ c o s θ c o s θ − 1 c o s θ + s i n θ c o s θ = t a n θ + s e c θ − 1 1 − s e c θ + t a n θ ⇒ ( t a n θ + s e c θ ) − ( s e c 2 θ − t a n 2 θ ) 1 − s e c θ + t a n θ . . . . ( 1 + tan 2 θ = sec 2 θ ) = ( tan θ + sec θ ) − ( sec θ − tan θ ) ( sec θ + tan θ ) 1 − sec θ + tan θ = ( tan θ + sec θ ) ( 1 − sec θ + tan θ ) ( 1 − sec θ + tan θ ) = tan θ + sec θ = sin θ cos θ + 1 cos θ = sin θ + 1 cos θ [Multiplying the numerator and denominator by ( 1 − sin θ ) ] = 1 + sin θ cos θ × 1 − sin θ 1 − sin θ = 1 − sin 2 θ cos θ ( 1 − sin θ ) = cos 2 θ cos θ ( 1 − sin θ ) = cos θ 1 − sin θ

Which of the following options is equal to sin θ+1−cos

Question

Which of the following options is equal to $\frac{sin θ + 1 - cos θ}{cos θ - 1 + sin θ} ?$

Options:

A .

\frac{cos θ}{1 - sin θ}

B .

\frac{1 + sin θ}{cos θ}

C .

\frac{1 + sin θ}{1 - sin θ}

D .

tan θ + sec θ

Answer: Option A
:
A, B, and D

$\frac{s i n θ + 1 - c o s θ}{c o s θ - 1 + s i n θ}$
Dividing the numerator and denominator by $c o s θ$ we get,
$\frac{\frac{s i n θ}{c o s θ} + \frac{1}{c o s θ} - \frac{c o s θ}{c o s θ}}{\frac{c o s θ}{c o s θ} - \frac{1}{c o s θ} + \frac{s i n θ}{c o s θ}}$
$= \frac{t a n θ + s e c θ - 1}{1 - s e c θ + t a n θ}$
$\Rightarrow \frac{(t a n θ + s e c θ) - (s e c^{2} θ - t a n^{2} θ)}{1 - s e c θ + t a n θ} . . . . (1 + {tan}^{2} θ = {sec}^{2} θ)$
$= \frac{(tan θ + sec θ) - (sec θ - tan θ) (sec θ + tan θ)}{1 - sec θ + tan θ}$
$= \frac{(tan θ + sec θ) (1 - sec θ + tan θ)}{(1 - sec θ + tan θ)}$
$= tan θ + sec θ$
$= \frac{sin θ}{cos θ} + \frac{1}{cos θ}$
$= \frac{sin θ + 1}{cos θ}$
[Multiplying the numerator and denominator by $(1 - sin θ)$ ]
$= \frac{1 + sin θ}{cos θ} \times \frac{1 - sin θ}{1 - sin θ}$
$= \frac{1 - {sin}^{2} θ}{cos θ (1 - sin θ)}$
$= \frac{{cos}^{2} θ}{cos θ (1 - sin θ)}$
$= \frac{cos θ}{1 - sin θ}$