Question

if cosec2θ (1 + cos θ ) (1 – cos θ) = k, then find the value of k.​

Answer 1

Given :

cosec2θ (1 + cos θ ) (1 – cos θ) = k

To find : k

Identity used :

• (a-b)(a+b) = (a² - b²)

• 1 - cos²θ = sin²θ

• cosecθ = 1 ÷ sinθ

• sin2θ = 2× cosθ × sinθ

• sinθ ÷ cosθ = tanθ

Solution :

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

ANSWER :

k = (1/2) × tanθ