Question

{1+cotA-cosecA} {1+tanA+secA}=2

Answer 1

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

And=2

Hope you get it :)

Answer 2

Step-by-step explanation:

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

And=2