Question

7 question pls 2 part​

Answer 1

Answer:

Sinθ(1 + Tanθ) + Cosθ(1 + Cotθ) = Secθ  + Cosecθ

Step-by-step explanation:

Sin theta( 1+tan theta) + cos theta(1+ cot theta) = sec theta + cosec theta

Sinθ(1 + Tanθ) + Cosθ(1 + Cotθ) = Secθ  + Cosecθ

LHS = Sinθ(1 + Tanθ) + Cosθ(1 + Cotθ)

Using Tanθ = Sinθ/Cosθ  & Cotθ = Cosθ/Sinθ

= Sinθ(1 + Sinθ/Cosθ) + Cosθ(1 + Cosθ/Sinθ)

=  (Sinθ/Cosθ)(Cosθ + Sinθ)  + (Cosθ/Sinθ)(Sinθ +  Cosθ)

= (Cosθ + Sinθ) (Sinθ/Cosθ + Cosθ/Sinθ)

=  (Cosθ + Sinθ)((Sin²θ + Cos²θ)/CosθSinθ)

Using Sin²θ + Cos²θ = 1

= (Cosθ + Sinθ)/CosθSinθ

= Cosθ/CosθSinθ +  Sinθ/CosθSinθ

= 1/Sinθ  +  1/Cosθ

= Cosecθ  + Secθ

= Secθ + Cosecθ

= RHS

QED

Proved

Sinθ(1 + Tanθ) + Cosθ(1 + Cotθ) = Secθ  + Cosecθ