Question

Tan theeta/1-cot there + cot theeta/ 1-tan theeta = 1+sec theeta . Cosec theeta​

Answer 1

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$\small\bold{\underline{\sf{\pink{Solution!!!}}}}$

$\sf \dfrac{tan\theta}{1-cot\theta}+\dfrac{cot\theta}{1-tan\theta}$

LHS :-

$\sf \dfrac{tan\theta}{1-cot\theta}+\dfrac{cot\theta}{1-tan\theta}$

$\sf tan\theta= \dfrac{sin\theta}{cos\theta}$

$\sf cot\theta= \dfrac{cos\theta}{sin\theta}$

$\sf \dfrac{\bigg(\dfrac{sin\theta}{cos\theta}\bigg)}{\bigg(1-\dfrac{cos\theta}{sin\theta}} \dfrac{\bigg(\dfrac{cos\theta}{sin\theta}\bigg)}{\bigg(1-\dfrac{sin\theta}{cos\theta}\bigg)}$

$\sf \dfrac{\bigg(\dfrac{sin\theta}{cos\theta}\bigg)}{\bigg(\dfrac{sin\theta-cos\theta}{sin\theta}\bigg)} \ + \dfrac{\bigg(\dfrac{cos\theta}{sin\theta}\bigg)}{\bigg(\dfrac{cos\theta-sin\theta}{cos\theta}\bigg)}$

$\sf \bigg(\dfrac{sin\theta}{cos\theta}\bigg)\times \bigg(\dfrac{sin\theta}{sin\theta-cos\theta}\bigg)\ + \bigg(\dfrac{cos\theta}{sin\theta}\bigg)\times \bigg(\dfrac{cos\theta}{cos\theta-sin\theta}\bigg)$

$\sf \dfrac{sin^2\theta}{cos\theta(sin\theta - cos\theta)} \ + \dfrac{cos^2\theta}{sin\theta(cos\theta-sin\theta)}$

$\sf \dfrac{sin^2\theta}{cos\theta(sin\theta - cos\theta)} \ - \dfrac{cos^2\theta}{sin\theta(sin\theta-cos\theta)}$

$\sf \dfrac{sin^3\theta-cos^3\theta}{sin\theta\ cos\theta(sin\theta-cos\theta)}$

${\sf {\small\ \ a^3-b^3= (a-b)(a^2+b^2+ab)}}$

Answer 2

Step-by-step explanation:

HELLO DEAR,

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

I HOPE IT'S HELP YOU DEAR,

THANKS