(sec² θ - 1)(1 - cosec² θ) is equal to (a) -1 (b) 1 (c) 0 (d) 2

Mathematics

(sec² θ - 1)(1 - cosec² θ) is equal to
-1
1
0
2

Trigonometric Identities

33 Likes

Answer

Given, (sec² θ - 1)(1 - cosec² θ)

By using trigonometric identities the equation can be written as,

$\Rightarrow \text{tan}^2 \text{ θ} (-\text{cot}^2 \text{ θ}) \\[1em] \Rightarrow \text{tan}^2 \text{ θ} \times \dfrac{-1}{\text{tan}^2 \text{ θ}} \\[1em] \Rightarrow -1.$

Hence, Option 1 is the correct option.

Answered By

3 Likes

Mathematics

(sec² θ - 1)(1 - cosec² θ) is equal to
-1
1
0
2

Trigonometric Identities

Answer

Related Questions

If x = h + a cos θ and y = k + a sin θ, prove that (x - h)² + (y - k)² = a².

$\text{cot}^2 \text{ θ} - \dfrac{1}{\text{sin}^2 \text{ θ}}$ is equal to
1
-1
sin² θ
sec² θ

$\dfrac{\text{tan}^2 \text{ θ}}{\text{1 + tan}^2 \text{ θ}}$ is equal to
2sin² θ
2cos² θ
sin² θ
cos² θ

(cos θ + sin θ)² + (cos θ - sin θ)² is equal to
-2
0
1
2

Mathematics

(sec2 θ - 1)(1 - cosec2 θ) is equal to-1102

Trigonometric Identities

Answer

Related Questions

If x = h + a cos θ and y = k + a sin θ, prove that (x - h)2 + (y - k)2 = a2.

cot2 θ−1sin2 θ\text{cot}^2 \text{ θ} - \dfrac{1}{\text{sin}^2 \text{ θ}}cot2 θ−sin2 θ1​ is equal to1-1sin2 θsec2 θ

tan2 θ1 + tan2 θ\dfrac{\text{tan}^2 \text{ θ}}{\text{1 + tan}^2 \text{ θ}}1 + tan2 θtan2 θ​ is equal to2sin2 θ2cos2 θsin2 θcos2 θ

(cos θ + sin θ)2 + (cos θ - sin θ)2 is equal to-2012

(sec² θ - 1)(1 - cosec² θ) is equal to
-1
1
0
2

If x = h + a cos θ and y = k + a sin θ, prove that (x - h)² + (y - k)² = a².

$\text{cot}^2 \text{ θ} - \dfrac{1}{\text{sin}^2 \text{ θ}}$ is equal to
1
-1
sin² θ
sec² θ

$\dfrac{\text{tan}^2 \text{ θ}}{\text{1 + tan}^2 \text{ θ}}$ is equal to
2sin² θ
2cos² θ
sin² θ
cos² θ

(cos θ + sin θ)² + (cos θ - sin θ)² is equal to
-2
0
1
2