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Answer:
Step-by-step explanation:
Given:
[tex]\tt \dfrac{tan\: \theta}{1+sec\: \theta} -\dfrac{tan\: \theta}{1-sec\: \theta} =2\:cosec\: \theta[/tex]
To Prove:
LHS = RHS
Proof:
First consider the LHS of the equation,
[tex]\tt \dfrac{tan\: \theta}{1+sec\: \theta} -\dfrac{tan\: \theta}{1-sec\: \theta}[/tex]
Cross multiplying we get,
[tex]\tt \implies \dfrac{tan\: \theta(1-sec\: \theta)-tan\: \theta(1+sec\: \theta)}{(1+sec\: \theta)(1-sec\: \theta)}[/tex]
We know that,
(a + b) (a - b) = a² - b²
Simplifying we get,
[tex]\tt \implies \dfrac{tan\: \theta-tan\: \theta sec\: \theta-tan\: \theta - tan\: \theta sec\: \theta}{1-sec^{2}\: \theta }[/tex]
We know,
sec² θ = 1 + tan²θ
1 - sec²θ = -tan²θ
[tex]\tt \implies \dfrac{-2tan\: \theta sec\: \theta}{-tan^{2}\: \theta }[/tex]
[tex]\tt \implies \dfrac{2\:sec\: \theta}{tan\: \theta}[/tex]
[tex]\tt \implies 2\times \dfrac{1}{cos\: \theta} \times \dfrac{cos\: \theta}{sin\: \theta}[/tex]
[tex]\tt \implies \dfrac{2}{sin\: \theta}[/tex]
We know that cosec θ = 1/sin θ
[tex]\tt \implies 2\: cosec\: \theta[/tex]
= RHS
Hence proved.