Question

prove that: (secA+ tanA)^2=
(Cosec A +1)/(CaseA-1)​

Answer 1

Step-by-step explanation:

(secA+tanA)^2

it is in the form of

(a+b)^2=a^2+b^2+2ab

sec^2A+tan^2A+2secAtanA

1/cos^2A+sin^2A/cos^2A+2sinA/cos^2A

(1+sin^2A+2sinA)/cos^2A

1+sin^2A+2sinA is in the form of (a+b)^2

(1+sinA)^2/(cos^2A)

(1+sinA)(1+sinA)/(1-sin^2A)

(1+sinA)(1+sinA)/(1+sinA)(1-sinA)

sinA can be written as 1/cosecA

(1+1/cosecA)/(1-1/cosecA

therefore (cosecA+1)/(cosecA-1)

(secA+tanA)^2=(cosecA+1)/(cosec-1)