Question

if sec A +tanA=p then prove that sin A =p square -1÷p square +1​

Answer 1

Answer:

Step-by-step explanation:

sec A + tan A = p   ........(Given)

therefore,

p^2 -1 = (Sec A + tan A)^2  - 1 .

therefore,

p^2 - 1 = (sec^2 A + tan^2 A + 2 sec A. tan A) - 1

          = sec^2 A - 1 + tan^2 A + 2 sec A. tan A

          = tan^2 A + tan^2 A + 2 sec A. tan A

         = 2 tan^2 A + 2 sec A. tan A

         = 2 tan A (tan A + sec A) .............(1)

p^2 + 1 = (sec^2 A + tan^2 A + 2 sec A. tan A) + 1

          = sec^2 A + 1 + tan^2 A + 2 sec A. tan A

          = sec^2 + sec^2  + 2 sec A. tan A

         = 2 sec^2  + 2 sec A. tan A

         = 2 sec A (tan A + sec A) .............(2)

therefore,

p^2 - 1     ===>      2 tan A (tan A + sec A)

p^2 + 1                   2 sec A (tan A + sec A)

                  = tan A / sec A

                 = sin A/ Cos A

                     1/ cos A

                 =  Sin A

Hence proved............!!!!