The correct option is D 2(1 + tan θ + sec θ ) (1 + cot θ cosec θ ) = (1+ sin θ/cos θ + 1/cos θ) (1+ cos θ/sin θ 1/sin θ) = (cos θ +sin θ +1)/cos θ×(sin θ ...

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Byju's Answer

Standard X

Mathematics

Trigonometric Ratios

1+tanθ+θ1+θ-c...

Question

$(1 + t a n θ + s e c θ) (1 + c o t θ - c o s e c θ)$

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Solution

The correct option is C 2
$(1 + t a n θ + s e c θ) (1 + c o t θ - c o s e c θ)$
$= (1 + \frac{s i n θ}{c o s θ} + \frac{1}{c o s θ}) (1 + \frac{c o s θ}{s i n θ} - \frac{1}{s i n θ})$
$= \frac{(c o s θ + s i n θ + 1)}{c o s θ} \times \frac{(s i n θ + c o s θ - 1)}{s i n θ}$
$= \frac{(c o s θ + s i n θ)^{2} - 1^{2}}{(c o s θ s i n θ)}$
$= \frac{(c o s^{2} θ + s i n^{2} θ + 2 c o s θ s i n θ - 1)}{(c o s θ s i n θ)}$
$= \frac{(1 + 2 c o s θ s i n θ - 1)}{(c o s θ s i n θ)}$
$= \frac{(2 c o s θ s i n θ)}{(c o s θ s i n θ)} = 2$

Suggest Corrections

(1+tanθ+secθ)(1+cotθ−cosecθ)

The correct option is C 2(1+tanθ+secθ)(1+cotθ−cosecθ) =(1+sinθcosθ+1cosθ)(1+cosθsinθ−1sinθ) =(cosθ+sinθ+1)cosθ×(sinθ+cosθ−1)sinθ =(cosθ+sinθ)2−12(cosθsinθ) =(cos2θ+sin2θ+2cosθsinθ−1)(cosθsinθ) =(1+2cosθsinθ−1)(cosθsinθ) =(2cosθsinθ)(cosθsinθ)=2

$(1 + t a n θ + s e c θ) (1 + c o t θ - c o s e c θ)$