Question

(1+cotA-cosecA) (1+tanA+secA)​

Answer 1

$\huge \purple {ANSWER}$

cosecA=1/sinA

secA=1/cosA

And cotA=(cosA/sinA)

tanA=(sinA/cosA)

So substituting value and taking LCM we get

(sinA+cosA-1)(cosA+sinA+1)/sinAcosA

Now (a+b)(a-b)=a^2-b^2

So,

((sinA+cosA)^2–1)/sinAcosA

(sin^2A+2sinAcosA+cos^2A-1)/sinAcosA

Now sin^2A+cos^2A = 1 SO….

2sinAcosA/sinAcosA = 2

Ans = 2