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

1.	Prove that: (secA+ tanA)^2= (Cosec A +1)/(CaseA-1)
Answer» 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)

Answer»

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)

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

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