Prove that (1 + cot A - cosec A ) (1 + tan A + sec A) = 2

#### Solution

We have to prove (1 + cot A - cosec A ) (1 + tan A + sec A) = 2

We know that, `sin^2 "A" + cos^2 "A" = 1.`

So,

`(1 + cot"A" - "cosec""A")(1 + tan "A"+ sec "A") = (1 + (cos"A")/(sin"A") - (1)/(sin"A"))(1+(sin"A")/(cos "A")+(1)/(cos"A"))`

= `((sin"A" + cos "A"-1)/(sin"A"))((cos"A" + sin"A"+1)/(cos"A"))`

= `((sin "A" + cos"A" - 1)(sin "A"+ cos"A" + 1))/(sin"A" cos"A")`

= `{{(sin "A" + cos "A")-1} {(sin"A" + cos"A")+1}}/(sin"A" cos"A")`

= `((sin "A" + cos "A")^2-1)/(sin"A" cos"A")`

= `(sin^2 "A" + 2sin "A" cos"A"+ cos^2 "A"-1)/(sin"A" cos "A")`

= `((sin^2 "A" + cos^2 "A") + 2sin "A" cos "A" - 1)/(sin "A" cos"A")`

= `(1 + 2sin "A" cos "A" - 1)/(sin "A" cos "A")`

= `(2sin "A" cos "A")/(sin "A" cos "A")`

=2

Hence proved.