Question

Prove that,tantheta/sectheta-1+tantheta/sectheta+1=2cosectheta​

Answer 1

We have to prove that,

$\sf \Longrightarrow \dfrac{tan \theta}{sec \theta - 1} + \dfrac{tan \theta}{sec \theta + 1} = 2 cosec \theta$

Solution:

Taking the LHS we get:

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

Take the LCM.

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

On applying the identity (a + b)(a - b) = a² - b² and on taking tanθ as common out from the numerator we get:

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

On applying the identity sec²θ - 1 = tan²θ we get:

$\sf \Longrightarrow \dfrac{tan \theta \Big(sec \theta + sec \theta\Big) }{tan^2 \theta}$

Cancelling tanθ and tan²θ we get:

$\sf \Longrightarrow \dfrac{2sec \theta}{tan \theta}$

We know that secθ = 1/cosθ and tanθ = sinθ/cosθ.

Substituting these values we get:

$\sf \Longrightarrow \dfrac{2 \times \bigg(\dfrac{1}{cos \theta}\bigg)}{\bigg(\dfrac{sin \theta}{cos \theta}\bigg)}$

$\sf \Longrightarrow \dfrac{2}{cos \theta} \ \times \ \dfrac{cos \theta}{sin \theta}$

$\sf \Longrightarrow \dfrac{2}{sin \theta}$

We know that sinθ = 1/cosecθ. Applying this relation we get:

$\sf \Longrightarrow 2 cosec \theta$

LHS = RHS

Hence proved.

Answer 2

sec²θ-1=tan²θ

(secθ-1)(secθ+1)=tanθ×tanθ

∴tanθ/(secθ-1)=(secθ+1)/tanθ 【Qed】

②tanθ/(secθ-1)=(secθ+1)/tanθ

LHS=(s/c)/{(1/c)-1}=(s/c)/(1-c/c)=s/(1-c)=s(1+c)/(1-c²)=s(1+c)/s²=(1+c)/s

RHS=((1/c)+1}(s/c)=c(1+c)/(sc)=(1+c)/s

LHS=RHS (PROVEN)

tanθsecθ−1=secθ+1tanθ

crossmultiplyingweget,

tan2θ=sec2θ−1

tan2θ=tan2θ[sec2θ−1=tan2θ]