(1 + cot 2 A) + (1 + tan 2 A) is equal to : 1. 2. sec 2 A - cosec 2 A 3. sin 2 A - sin 4 A 4. sec 2 A.cosec 2 A

Solving, ⇒ (1 + cot 2 A) + (1 + tan 2 A) ⇒ cosec 2 A + sec 2 A ⇒ ⇒ Substituting, sin 2 A + cos 2 A = 1, we get : ⇒ ⇒ cosec 2 A. sec 2 A Hence, Option 4 is the correct option.

Mathematics

(1 + cot² A) + (1 + tan² A) is equal to :
$\dfrac{1}{\text{sin}^2 A - \text{cos}^2 A}$
sec² A - cosec² A
sin² A - sin⁴ A
sec² A.cosec² A

Trigonometric Identities

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Answer

Solving,

⇒ (1 + cot² A) + (1 + tan² A)

⇒ cosec² A + sec² A

⇒ $\dfrac{1}{\text{sin}^2 A} + \dfrac{1}{\text{cos}^2 A}$

⇒ $\dfrac{\text{cos}^2 A + \text{sin}^2 A}{\text{sin}^2 A. \text{cos}^2 A}$

Substituting, sin² A + cos² A = 1, we get :

⇒ $\dfrac{1}{\text{sin}^2 A. \text{cos}^2 A}$

⇒ cosec² A. sec² A

Hence, Option 4 is the correct option.

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Mathematics

(1 + cot² A) + (1 + tan² A) is equal to :
$\dfrac{1}{\text{sin}^2 A - \text{cos}^2 A}$
sec² A - cosec² A
sin² A - sin⁴ A
sec² A.cosec² A

Trigonometric Identities

Answer

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$\dfrac{\text{cos θ cot θ}}{\text{1 + sin θ}}$ = cosec θ - 1

If a = tan θ and b = sec θ, the relation between a and b is :
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$\dfrac{1}{\text{cot θ + tan θ}}$ is equal to :
sin θ + cos θ
sin θ. cos θ
$\dfrac{1}{\text{sin θ. cos θ}}$
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Mathematics

(1 + cot2 A) + (1 + tan2 A) is equal to :1sin2A−cos2A\dfrac{1}{\text{sin}^2 A - \text{cos}^2 A}sin2A−cos2A1​sec2 A - cosec2 Asin2 A - sin4 Asec2 A.cosec2 A

Trigonometric Identities

Answer

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Prove the following identities :1 + sin Acosec A - cot A−1 - sin Acosec A + cot A\dfrac{\text{1 + sin A}}{\text{cosec A - cot A}} - \dfrac{\text{1 - sin A}}{\text{cosec A + cot A}}cosec A - cot A1 + sin A​−cosec A + cot A1 - sin A​ = 2(1 + cot A)

Prove the following identities :cos θ cot θ1 + sin θ\dfrac{\text{cos θ cot θ}}{\text{1 + sin θ}}1 + sin θcos θ cot θ​ = cosec θ - 1

If a = tan θ and b = sec θ, the relation between a and b is :a × b = 1a2 - b2 = 1b2 - a2 = 1a2 + b2 = 1

1cot θ + tan θ\dfrac{1}{\text{cot θ + tan θ}}cot θ + tan θ1​ is equal to :sin θ + cos θsin θ. cos θ1sin θ. cos θ\dfrac{1}{\text{sin θ. cos θ}}sin θ. cos θ1​1sin θ + cos θ\dfrac{1}{\text{sin θ + cos θ}}sin θ + cos θ1​

(1 + cot² A) + (1 + tan² A) is equal to :
$\dfrac{1}{\text{sin}^2 A - \text{cos}^2 A}$
sec² A - cosec² A
sin² A - sin⁴ A
sec² A.cosec² A

Prove the following identities :
$\dfrac{\text{1 + sin A}}{\text{cosec A - cot A}} - \dfrac{\text{1 - sin A}}{\text{cosec A + cot A}}$ = 2(1 + cot A)

Prove the following identities :
$\dfrac{\text{cos θ cot θ}}{\text{1 + sin θ}}$ = cosec θ - 1

If a = tan θ and b = sec θ, the relation between a and b is :
a × b = 1
a² - b² = 1
b² - a² = 1
a² + b² = 1

$\dfrac{1}{\text{cot θ + tan θ}}$ is equal to :
sin θ + cos θ
sin θ. cos θ
$\dfrac{1}{\text{sin θ. cos θ}}$
$\dfrac{1}{\text{sin θ + cos θ}}$