Question

Hii guys

the 2 ND one

Answer 1

$\textbf{Answer}$

We $\textbf{will use following identities}$ in the solution,

•sin^2 A + cos^2 A = 1
•cot A = cos A / sin A
•tan A = sin A / cos A
•a^2 - b^2 = (a+b)(a-b)

$\textbf{We need to prove that,}$

(1 + cot θ - cosec θ) (1 + tan θ + sec θ) = 2

=> LHS = (1 + cot θ - cosec θ) (1 + tan θ + sec θ)

=> LHS = (1 + cosθ/sinθ - 1/sinθ)(1 + sinθ/cosθ + 1/cosθ)

=> LHS = (sinθ + cosθ - 1)(sinθ + cosθ + 1) / (sinθcosθ)

=>LHS = {(sinθ + cos)^2 - 1^2} / (sinθcosθ)

=> LHS = (sin^2θ + cos^2θ + 2sinθcosθ - 1)/(sinθcosθ)

=> LHS = (1 - 1 + 2sinθcosθ)/(sinθcosθ)

=> $\textbf{LHS = 2 = RHS}$

$\textbf{Hope My Answer Helped}$
$\textbf{Thanks}$