Question

Prove the following trigonometric identities.
(i)$cot\Theta-tan\Theta=\frac{2cos^{2}\Theta-1 }{sin\Theta cos\Theta}$
(ii)$tan\Theta-cot\Theta=\frac{2sin^{2}\Theta-1 }{sin\Theta cos\Theta}$

Answer 1

Answer with Step-by-step explanation:

Given : (i) cot θ – tan θ = 2cos²θ −1/(sinθ × cosθ)

L.H.S = cot θ – tan θ

= cosθ/sinθ – sinθ/cosθ

[By using the identity, cotθ = cosθ/sinθ  ,  tanθ = sinθ/cosθ ]

= (cos²θ − sin²θ)/sinθ×cosθ

[By taking LCM]

= cos²θ − (1− cos²θ)sinθ×cosθ

[By using the identity, sin²θ = (1- cos²θ)]

= (cos²θ −1 + cos²θ) /(sinθ×cosθ)

= (2cos²θ − 1)/ (sinθ × cosθ)

cot θ – tan θ = 2cos²θ −1/(sinθ × cosθ)

L.H.S = R.H.S  

Hence Proved..

(ii) Given : tanθ− cotθ = (2sin²θ −1)/sinθ × cosθ)

L.H.S = tanθ− cotθ

= sinθ/cosθ  – cosθ/sinθ

[By using the identity, cotθ = cosθ/sinθ  ,  tanθ = sinθ/cosθ ]

= (sin²θ − cos²θ)/(sinθ × cosθ)

[By taking LCM]

= sin²θ − (1− sin²θ)/(sinθ × cosθ)

[By using the identity , cos² θ = 1 - sin² θ]

= sin²θ −1 + sin²θ /(sinθ × cosθ)

= (2sin²θ − 1)/ (sinθ × cosθ)

tanθ− cotθ = (2sin²θ −1)/sinθ × cosθ)

L.H.S = R.H.S  

Hence Proved..

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Answer 2

Answer:

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