The correct option is C 1+cos θ/1 cos θ tan^2 θ/(sec θ 1)^2 = sec^2 θ 1/(sec θ 1)^2 =(sec θ 1)(sec θ +1)/(sec θ 1)(sec θ 1) =1/cos θ+1/1/cos θ 1 =1+cos θ/cos θ ...

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Byju's Answer

Standard X

Mathematics

Trigonometric Identities

tan 2θ/θ-12=

Question

$\frac{{tan}^{2} θ}{(sec θ - 1)^{2}} =$

$\frac{1 + sin θ}{1 - sin θ}$

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$1$

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$tan θ$

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$\frac{1 + cos θ}{1 - cos θ}$

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Solution

The correct option is D
$\frac{1 + cos θ}{1 - cos θ}$

$\frac{{tan}^{2} θ}{(sec θ - 1)^{2}} = \frac{{sec}^{2} θ - 1}{(sec θ - 1)^{2}}$

$= \frac{(sec θ - 1) (sec θ + 1)}{(s e c θ - 1) (sec θ - 1)}$

$= \frac{\frac{1}{cos θ} + 1}{\frac{1}{cos θ} - 1}$

$= \frac{\frac{1 + cos θ}{cos θ}}{\frac{1 - cos θ}{cos θ}}$

$= \frac{1 + cos θ}{1 - cos θ}$

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Similar questions

Q. Prove that

\frac{1 + cos θ}{1 - cos θ} = \frac{{tan}^{2} θ}{{(sec θ - 1)}^{2}}

Q. If

cot θ = \frac{12}{5}

show that

{tan}^{2} θ - {sin}^{2} θ = {sin}^{4} θ \times {sec}^{2} θ

Q. If

θ = 30^{\circ}

, verify that :
(i)

tan 2 θ = \frac{2 tan θ}{1 - {tan}^{2} θ}

(ii)

tan 2 θ = \frac{4 tan θ}{1 + {tan}^{2} θ}

(iii)

cos 2 θ = \frac{1 - {tan}^{2} θ}{1 + {tan}^{2} θ}

(iv)

cos 3 θ = 4 {cos}^{3} θ - 3 cos θ

(1 + s e c θ) (1 - s e c θ) + t a n^{2} θ =

Q. Evaluate:

{c o s e c}^{2} (90 - θ) - {t a n}^{2} θ

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tan2 θ(sec θ−1)2=

The correct option is D 1+cos θ1−cos θ tan2 θ(sec θ−1)2=sec2 θ−1(sec θ−1)2 =(sec θ−1)(sec θ+1)(sec θ−1)(sec θ−1) =1cos θ+11cos θ−1 =1+cos θcos θ1−cos θcos θ =1+cos θ1−cos θ

$\frac{{tan}^{2} θ}{(sec θ - 1)^{2}} =$