The numerical value of cosec-1-cos-sin-sin-1-cos-2 2- is

The correct option is A 2cosec θ[1 cosθsinθ+sinθ1 cosθ] 2^2θ = nbsp;[1 cosθsin^2 θ+11 cosθ] 2^2θ = nbsp;[1 cosθ1 cos^2θ+11 cosθ] 2^2θ = nbsp;[11+cosθ+ ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Common Angles

The numerical...

Question

The numerical value of $cosec θ [\frac{1 - cos θ}{sin θ} + \frac{sin θ}{1 - cos θ}] - 2 {cot}^{2} θ$ is

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Solution

The correct option is C $2$
$cosec θ [\frac{1 - cos θ}{sin θ} + \frac{sin θ}{1 - cos θ}] - 2 {cot}^{2} θ = [\frac{1 - cos θ}{{sin}^{2} θ} + \frac{1}{1 - cos θ}] - 2 {cot}^{2} θ = [\frac{1 - cos θ}{1 - {cos}^{2} θ} + \frac{1}{1 - cos θ}] - 2 {cot}^{2} θ = [\frac{1}{1 + cos θ} + \frac{1}{1 - cos θ}] - 2 {cot}^{2} θ = [\frac{1 + cos θ + 1 - cos θ}{1 - {cos}^{2} θ}] - 2 {cot}^{2} θ = 2 {cosec}^{2} θ - 2 {cot}^{2} θ = 2$

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Similar questions

Q. Is LHS=RHS?

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

Q. Show that

\frac{sin θ + cos θ}{sin θ - cos θ} + \frac{sin θ - cos θ}{sin θ + cos θ}

= - 2 sec 2 θ

Prove the following trigonometric identities.
(i) $\frac{1 + cos θ + sin θ}{1 + cos θ - sin θ} = \frac{1 + sin θ}{cos θ}$

(ii) $\frac{sin θ - cos θ + 1}{sin θ + cos θ - 1} = \frac{1}{sec θ - tan θ}$

(iii) $\frac{cos θ - sin θ + 1}{cos θ + sin θ - 1} = cosec θ + cot θ$

(iv) $(sin θ + cos θ) (tan θ + cot θ) = sec θ + cosec θ$

Q. Solve the following equations:
(i)

\sin θ + \cos θ = \sqrt{2}

(ii)

\sqrt{3} \cos θ + \sin θ = 1

(iii)

\sin θ + \cos θ = 1

(iv)

cosec θ = 1 + \cot θ

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The numerical value of cosec θ[1−cosθsinθ+sinθ1−cosθ]−2cot2θ is

The correct option is C 2cosec θ[1−cosθsinθ+sinθ1−cosθ]−2cot2θ=[1−cosθsin2θ+11−cosθ]−2cot2θ=[1−cosθ1−cos2θ+11−cosθ]−2cot2θ=[11+cosθ+11−cosθ]−2cot2θ=[1+cosθ+1−cosθ1−cos2θ]−2cot2θ=2 cosec 2θ−2cot2θ=2

The numerical value of $cosec θ [\frac{1 - cos θ}{sin θ} + \frac{sin θ}{1 - cos θ}] - 2 {cot}^{2} θ$ is