Question

25. If tan’e (cosec 0 - 1) (cosec 0 + 1) = k, find the value of k.
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Answer 1

Required Answer:-

Given:

• tan²θ(cosec θ - 1)(cosec θ + 1) = k

To Find:

• The value of k.

Solution:

Given,

→ tan²θ(cosec θ - 1)(cosec θ + 1) = k

Using identity (a + b)(a - b) = a² - b², we get,

→ tan²θ × (cosec²θ - 1) = k

We know that,

→ cosec²θ - cot²θ = 1

→ cosec²θ - 1 = cot²θ

Substituting the value in the equation, we get,

→ tan²θ × cot²θ = k

As tan θ is the reverse of cot θ i.e.,

→ tan θ = 1/cot θ

→ 1/cot²θ × cot²θ = k

→ 1 = k

→ k = 1

→ So, the value of k is 1.

Answer:

• k = 1

Additional Formulae:

1. Relationship between sides and T-Ratios.

• sin θ = Height/Hypotenuse.
• cos θ = Base/Hypotenuse.
• tan θ = Height/Base.
• cot θ = Base/Height.
• sec θ = Hypotenuse/Base.
• cosec θ = Hypotenuse/Height.

2. Reciprocal Identities.

• sin θ = 1/cosec θ and cosec θ = 1/sin θ
• cos θ = 1/sec θ and sec θ = 1/cot θ
• tan θ = 1/cot θ and cot θ = 1/tan θ

3. Co-function identities.

• sin(90° - θ) = cos θ and cos(90° - θ) = sin θ.
• tan(90° - θ) = cot θ and cot(90° - θ) = tan θ.
• sec(90° - θ) = cosec θ and cosec(90° - θ) = sec θ

4. Pythagoras identities.

• sin²θ + cos²θ = 1
• cosec²θ - cot²θ = 1
• sec²θ - tan²θ = 1

Answer 2

Given :-

$\sf {tan}^{2} \theta(cosec \theta - 1)(cosec \theta + 1) = k$

To Find :-

Value of k.

Solution :-

$: \implies \sf {tan}^{2} \theta(cosec \theta - 1)(cosec \theta + 1) = k$

$: \implies \sf {tan}^{2} \theta \times( {cosec}^{2} \theta - {1}^{2} ) = k$

$\bf\dag \: tan \theta = \dfrac{sin \theta}{cos \theta}$

$\bf\dag \: cosec \theta = \dfrac{1}{sin \theta}$

$: \implies \sf \dfrac{ {sin}^{2} \theta}{ {cos}^{2} \theta} \times \left( \dfrac{1}{ {sin}^{2} \theta} - 1\right) = k$

$: \implies \sf \dfrac{ {sin}^{2} \theta}{ {cos}^{2} \theta} \times \dfrac{1 - {sin}^{2} \theta}{ {sin}^{2} \theta} = k$

$\bf \dag \: 1 - {sin}^{2} \theta = {cos}^{2} \theta$

$: \implies \sf \dfrac{ {sin}^{2} \theta}{ {cos}^{2} \theta} \times \dfrac{ {cos}^{2} \theta}{ {sin}^{2} \theta} = k$

$: \implies \sf 1 = k$

$\underline{ \bf \: k = 1}$