Question

tan theta upon 1-cot theta +cot theta upon 1-tan theta =1+sec theta cosec theta​

Answer 1

GIVEN:- tanθ/(1 - cotθ) + cotθ/(1 - tanθ)

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

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

=> 1/(tanθ - 1) { tan²θ - 1/tanθ }

=> 1/(tanθ - 1) { (tan³θ - 1)/tanθ)

[as, a³ - b³ = (a - b)(a² + b² + ab)

=> {(tanθ - 1)(tan²θ + 1 + tanθ)}/{(tanθ - 1)(tanθ)}

=> tanθ + cotθ + 1

=> sinθ/cosθ + cosθ/sinθ + 1

=> (sin²θ + cos²θ)/sinθ . cosθ + 1

=> 1/sinθ . cosθ + 1

=> cosecθ . secθ + 1

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Answer 2

$\small{\underline{\underline{\sf {Given:}}}$

tanθ/(1 - cotθ) + cotθ/(1 - tanθ)

$\small\underlined{\underline{\sf{Solution:}}}$

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

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

=> 1/(tanθ - 1) { tan²θ - 1/tanθ }

=> 1/(tanθ - 1) { (tan³θ - 1)/tanθ)

[as, a³ - b³ = (a - b)(a² + b² + ab)

=> {(tanθ - 1)(tan²θ + 1 + tanθ)}/{(tanθ - 1)(tanθ)}

=> tanθ + cotθ + 1

=> sinθ/cosθ + cosθ/sinθ + 1

=> (sin²θ + cos²θ)/sinθ . cosθ + 1

=> 1/sinθ . cosθ + 1

=> cosecθ . secθ + 1