Question

prove that the following question​

Answer 1

Answer:

Step-by-step explanation:

Given:

$\tt \dfrac{tan\: \theta}{1+sec\: \theta} -\dfrac{tan\: \theta}{1-sec\: \theta} =2\:cosec\: \theta$

To Prove:

LHS = RHS

Proof:

First consider the LHS of the equation,

$\tt \dfrac{tan\: \theta}{1+sec\: \theta} -\dfrac{tan\: \theta}{1-sec\: \theta}$

Cross multiplying we get,

$\tt \implies \dfrac{tan\: \theta(1-sec\: \theta)-tan\: \theta(1+sec\: \theta)}{(1+sec\: \theta)(1-sec\: \theta)}$

We know that,

(a + b) (a - b) = a² - b²

Simplifying we get,

$\tt \implies \dfrac{tan\: \theta-tan\: \theta sec\: \theta-tan\: \theta - tan\: \theta sec\: \theta}{1-sec^{2}\: \theta }$

We know,

sec² θ = 1 + tan²θ

1 - sec²θ = -tan²θ

$\tt \implies \dfrac{-2tan\: \theta sec\: \theta}{-tan^{2}\: \theta }$

$\tt \implies \dfrac{2\:sec\: \theta}{tan\: \theta}$

$\tt \implies 2\times \dfrac{1}{cos\: \theta} \times \dfrac{cos\: \theta}{sin\: \theta}$

$\tt \implies \dfrac{2}{sin\: \theta}$

We know that cosec θ = 1/sin θ

$\tt \implies 2\: cosec\: \theta$

= RHS

Hence proved.