If cosec titar+ cot titar=K then prove that cos titar=ke power 2 -1 by

1.	If cosec titar+ cot titar=K then prove that cos titar=ke power 2 -1 bye ke power 2 +1
Answer» Answer:- GIVEN: Cosec θ + cot θ = k -- equation (1) we know that, cosec² θ - cot² θ = 1 a² - b² = (a + B)(a - b) So, ⟶ (cosec θ + cot θ) (cosec θ - cot θ) = 1 ⟶ k * (cosec θ - cot θ) = 1 [ From equation (1) ] ⟶ Cosec θ - cot θ = 1/k -- equation (2) Add equations (1) and (2) $\longrightarrow \sf \: \csc \theta + \cot \theta + \csc \theta - \cot \theta = k + \frac{1}{k} \\ \\ \\\longrightarrow \sf \: 2 \csc \theta = \frac{ {k}^{2} + 1 }{k} \\ \\ \\ \longrightarrow \sf\csc \theta = \frac{ {k}^{2} + 1 }{k} \times \frac{1}{2} \\ \\ \\ \longrightarrow \boxed{\sf\csc \theta = \frac{ {k}^{2} + 1 }{2k} }$ cot θ = Cosec θ * COS θ So substitute the value of cosec θ in equation (1). ⟶ Cosec θ + cosec θ * cos θ = k ⟶ cosec θ (1 + cos θ) = k $\: \longrightarrow \sf \: \frac{ {k}^{2} + 1 }{2k} \bigg(1 + \cos \theta \bigg) = k \\ \\ \\ \longrightarrow \sf \:1 + \cos \theta =k \times \frac{2k}{ {k}^{2} + 1} \\ \\\\ \longrightarrow \sf \: \cos \theta = \frac{2 {k}^{2} }{ {k}^{2} + 1 } - 1 \\ \\\\ \longrightarrow \sf \:\cos \theta = \frac{2 {k}^{2} - ( {k}^{2} + 1)}{ {k}^{2} + 1} \\\\ \\ \longrightarrow \sf \:\cos \theta = \frac{2 {k}^{2} - {k}^{2} - 1}{ {k}^{2} + 1} \\\\ \\ \longrightarrow \boxed{\sf \:\cos \theta = \frac{{k}^{2} - 1}{ {k}^{2} + 1}}$ Hence Proved.

If cosec titar+ cot titar=K then prove that cos titar=ke power 2 -1 bye ke power 2 +1

Answer»

Answer:-

GIVEN:

Cosec θ + cot θ = k -- equation (1)

we know that,

cosec² θ - cot² θ = 1

a² - b² = (a + B)(a - b)

So,

⟶ (cosec θ + cot θ) (cosec θ - cot θ) = 1

⟶ k * (cosec θ - cot θ) = 1

[ From equation (1) ]

⟶ Cosec θ - cot θ = 1/k -- equation (2)

Add equations (1) and (2)

$\longrightarrow \sf \: \csc \theta + \cot \theta + \csc \theta - \cot \theta = k + \frac{1}{k} \\ \\ \\\longrightarrow \sf \: 2 \csc \theta = \frac{ {k}^{2} + 1 }{k} \\ \\ \\ \longrightarrow \sf\csc \theta = \frac{ {k}^{2} + 1 }{k} \times \frac{1}{2} \\ \\ \\ \longrightarrow \boxed{\sf\csc \theta = \frac{ {k}^{2} + 1 }{2k} }$

cot θ = Cosec θ * COS θ

So substitute the value of cosec θ in equation (1).

⟶ Cosec θ + cosec θ * cos θ = k

⟶ cosec θ (1 + cos θ) = k

$\: \longrightarrow \sf \: \frac{ {k}^{2} + 1 }{2k} \bigg(1 + \cos \theta \bigg) = k \\ \\ \\ \longrightarrow \sf \:1 + \cos \theta =k \times \frac{2k}{ {k}^{2} + 1} \\ \\\\ \longrightarrow \sf \: \cos \theta = \frac{2 {k}^{2} }{ {k}^{2} + 1 } - 1 \\ \\\\ \longrightarrow \sf \:\cos \theta = \frac{2 {k}^{2} - ( {k}^{2} + 1)}{ {k}^{2} + 1} \\\\ \\ \longrightarrow \sf \:\cos \theta = \frac{2 {k}^{2} - {k}^{2} - 1}{ {k}^{2} + 1} \\\\ \\ \longrightarrow \boxed{\sf \:\cos \theta = \frac{{k}^{2} - 1}{ {k}^{2} + 1}}$

Hence Proved.