Question

(tan theta - cot theta) ^ +2 =sec^ theta +cosec^ theta -2​

Answer 1

Answer:

$\bf{(tan\theta-cot\theta)^2+2=sec^2\theta-+cosec^2\theta-2}$

Step-by-step explanation:

$(tan\theta-cot\theta)^2+2$

$=tan^2\theta+cot^2\theta-2\:tan\theta\:cot\theta+2$

$=(sec^2\theta-1)+(cosec^2\theta-1)-2\:tan\theta\:(\frac{1}{tan\theta})+2$

$=sec^2\theta-+cosec^2\theta-2-2+2$

$=sec^2\theta-+cosec^2\theta-2$

$\implies\:\bf{(tan\theta-cot\theta)^2+2=sec^2\theta-+cosec^2\theta-2}$

Answer 2

Answer:

(Tanθ - Cotθ)² + 2  = Sec²θ + Cosec²θ  - 2

Step-by-step explanation:

(tan theta - cot theta) ^ +2 =sec^ theta +cosec^ theta -2​

=> (Tanθ - Cotθ)² + 2  = Sec²θ + Cosec²θ  - 2

=>  Tan²θ + Cot²θ - 2TanθCotθ + 2 = Sec²θ + Cosec²θ  - 2

=> Tan²θ + Cot²θ - 2 + 2 = Sec²θ + Cosec²θ  - 2

=> Tan²θ + Cot²θ = Sec²θ + Cosec²θ  - 2

=> Sin²θ/Cos²θ +   Cos²θ/Sin²θ = Sec²θ + Cosec²θ  - 2

=>( 1 - Cos²θ)/Cos²θ + (1 - Sin²θ)/Sin²θ = Sec²θ + Cosec²θ  - 2

=> 1/Cos²θ - 1 + 1/Sin²θ - 1  =Sec²θ + Cosec²θ  - 2

=> Sec²θ  + Cosec²θ - 2 = Sec²θ + Cosec²θ  - 2

=> LHS = RHS

QED

Proved

(Tanθ - Cotθ)² + 2  = Sec²θ + Cosec²θ  - 2