Question

prove this please?
could you please do it faster​

Answer 1

Answer:

LHS = RHS

Step-by-step explanation:

To Prove:

$\Longrightarrow \sf \dfrac{sin\theta -cos\theta+1}{sin\theta+cos\theta-1} = \dfrac{1}{sec\theta - tan\theta}$

$\sf LHS = \dfrac{sin\theta -cos\theta+1}{sin\theta+cos\theta-1}$

Dividing the numerator and denominator by cosθ we get:

(We're dividing it by cosθ to express sinθ and cosθ in terms of secθ and tanθ.)

$\sf LHS = \dfrac{\left(\dfrac{sin\theta}{cos\theta}-\dfrac{cos\theta}{cos\theta}+\dfrac{1}{cos\theta}\right)}{\left(\dfrac{sin\theta}{cos\theta}+\dfrac{cos\theta}{cos\theta}-\dfrac{1}{cos\theta}\right)}$

We know that.

⇒ sinθ/cosθ = tanθ

⇒ 1/cosθ = secθ

Applying these above we get,

$\sf LHS = \dfrac{tan\theta-1+sec\theta}{tan\theta+1-sec\theta}$

We know that sec²θ - tan²θ = 1

Substituting this in the place of 1 in the numerator we get,

$\sf LHS = \dfrac{tan\theta+sec\theta-(sec^2\theta-tan^2\theta)}{tan\theta+1-sec\theta}$

$\sf LHS = \dfrac{tan\theta+sec\theta-(sec\theta+tan\theta)(sec\theta-tan\theta)}{tan\theta+1-sec\theta}$

Taking tanθ + secθ as common from the numerator we get,

$\sf LHS = \dfrac{tan\theta+sec\theta(1-(sec\theta-tan\theta))}{tan\theta+1-sec\theta}$

$\sf LHS = \dfrac{tan\theta+sec\theta(1-sec\theta+tan\theta)}{tan\theta+1-sec\theta}$

$\sf LHS = \dfrac{tan\theta+sec\theta(tan\theta+1-sec\theta)}{tan\theta+1-sec\theta}$

Tanθ + 1 - Secθ gets cancelled as it's common both in the numerator and denominator.

$\sf LHS = tan\theta+sec\theta$

Now, dividing the numerator and denominator by (secθ - tanθ) we get,

$\sf LHS = \dfrac{tan\theta+sec\theta(sec\theta-tan\theta)}{sec\theta-tan\theta}$

$\sf LHS = \dfrac{sec^2\theta-tan^2\theta}{sec\theta-tan\theta}$

We know that,

sec²θ - tan²θ = 1

$\sf LHS = \dfrac{1}{sec\theta-tan\theta}$

LHS = RHS

Hence Proved.