Question

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

Answer 1

Answer:

Step-by-step explanation:

Given : tanθ/(1 − cotθ) + cotθ/(1 − tanθ) = 1 + tanθ + cotθ

LHS :  tanθ/(1− cotθ) + cotθ/(1− tanθ)

= tanθ/(1 - 1/tanθ) + (1/tanθ)/(1 - tanθ)

[By using the identity, cotθ = 1/tanθ ,  tanθ = 1/cotθ ]

= tanθ/[(tanθ - 1)/tanθ] + (1/tanθ)/(1 - tanθ)

= tanθ × tanθ / (tanθ - 1) + (1/tanθ)/(1 - tanθ)

= tan²θ/(tanθ - 1) + 1/tanθ(1 - tanθ)

= tan²θ/(tanθ - 1) - 1/tanθ(tanθ - 1)  

= (tan³θ - 1)/[tanθ(tanθ - 1)]

[By taking LCM]

= (tanθ - 1)(tan²θ + 1 + tanθ)/[tanθ(tanθ - 1)]  

[By using identity , a³ - b³ = (a - b) (a² + ab + b²)]

= (tan²θ + 1 + tanθ)/tanθ  

= tan²θ /tanθ + 1/tanθ + tanθ/tanθ  

= tanθ + cotθ + 1  

= 1 + tanθ + cotθ

tanθ/(1− cotθ) + cotθ/(1− tanθ) = 1 + tanθ + cotθ

L.H.S = R.H.S  

Hence Proved..

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Answer 2

According to the Question:

» tanθ/(1 − cotθ) + cotθ/(1 − tanθ) = 1 + tanθ + cotθ

Where,

LHS : tanθ/(1− cotθ) + cotθ/(1− tanθ)

» tanθ/(1 - 1/tanθ) + (1/tanθ)/(1 - tanθ)

Note: Using identity - [ cotθ = 1/tanθ , tanθ = 1/cotθ ]

» tanθ/[(tanθ - 1)/tanθ] + (1/tanθ)/(1 - tanθ)

» tanθ × tanθ / (tanθ - 1) + (1/tanθ)/(1 - tanθ)

» tan²θ/(tanθ - 1) + 1/tanθ(1 - tanθ)

» tan²θ/(tanθ - 1) - 1/tanθ(tanθ - 1)

» (tan³θ - 1)/[tanθ(tanθ - 1)]

Taking LCM, we get :-

» (tanθ - 1)(tan²θ + 1 + tanθ)/[tanθ(tanθ - 1)]

Note: Using identity - [ a³ - b³ = (a - b) (a² + ab + b²) ]

» (tan²θ + 1 + tanθ)/tanθ

» tan²θ /tanθ + 1/tanθ + tanθ/tanθ

» 1 + tanθ + cotθ

Also,

» tanθ/(1− cotθ) + cotθ/(1− tanθ) = 1 + tanθ + cotθ

∴ LHS = RHS _______[VERIFIED]