Question

prove that
(secA + cosecA)[(sinA + cosA)^2 - 1] = 2( sinA + cosA)​

Answer 1

Step-by-step explanation:

What is the answer to (secA+cosecA) (sinA+cosA) =2+secA.cosecA?

Super fast, super safe charging.

Chandra Kanth

Answered 2 years ago

Given :-

(secA+cosecA)(sinA+cosA)=2+secA.cosecA.

Consider LHS

(secA+cosecA)(sinA+cosA)

=secA(sinA+cosA)+cosecA(sinA+cosA)

=secA.sinA+secA.cosA+cosecA.sinA+cosecA.cosA

(As we know that secA.cosA=1 and cosecA.sinA=1)

=sinA.secA+1+1+cosecA.cosA

=sinA.1/cosA+2+cosA.1/sinA

hope it's helpful to you

=2+tanA+cotA

=2+sinA/cosA+cosA/sinA.

=2+(sin²A+cos²A)/cosA.sinA

(As we know that sin²A+cos²A=1)

=2+1/cosA.sinA

But wkt 1/sinA=cosecA and 1/cosceA

Therefore LHS=2+secA.cosecA.

Hence LHS=RHS