Question

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

Answer 1

Answer with Step-by-step explanation:

Given :  

(1 + tan²θ)/ (1+ cot²θ/ = [(1 − tanθ) /cotθ]² − tan²θ

LHS = (1 + tan²θ) / (1 + cot²θ)

= sec²θ/cosec²θ                                                        

[By using the identity, tan²θ + 1 = sec²θ , 1 + cot²θ = cosec²θ]

= sec²θ × 1 /cosec²θ

= 1/cos²θ × sin²θ

[By using the identity, secθ = 1/ cosθ & sin θ = 1/cosecθ]

= sin²θ/cos²θ

= tan²θ

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

(1 + tan²θ)/ (1+ cot²θ/ = [(1 − tanθ) /cotθ]² − tan²θ

L.H.S = R.H.S  

Hence Proved..

HOPE THIS ANSWER WILL HELP YOU...

Answer 2

Answer:

Refer to the attachment for your answer.