Question

sec A +tan A =p ,show that sinA=p×p-1÷p×p+1

Answer 1

If secA+tanA = p. Prove that sinA = p2-1/p2+1

Given p = secA+tanA

So, p2 - 1 = (secA+tanA)2 - 1

= (sec2A + 2.secA.tanA + tan2A) - 1 [Identity (a + b)2 = a2 +2ab +b2]

Since sec2A - 1 = tan2A:
sec2A + 2.secA.tanA + tan2A - 1 become tan2A + 2.secA.tanA + tan2A

= 2 tan2A + 2.secA.tanA

= 2 tanA (tanA + secA)

So, p2 - 1 = 2 tan (tanA + secA)

Similarly p2 +1 = (secA+tanA)2 + 1

 

= (sec2A + 2.secA.tanA + tan2A) + 1 [Identity (a + b)2 = a2 +2ab +b2]

Since tan2A + 1 = sec2A

sec2A + 2.secA.tanA + tan2A + 1 becomes sec2A + 2.secA.tanA + sec2A

= 2 sec2A + 2.secA.tanA

= 2 secA (secA + tanA)

So, p2 +1 = 2 secA (secA + tanA)

Now, (p2-1) / (p2 +1)

= 2 tanA (tanA + secA)
...2secA(secA+tanA) 

= 2tanA
...2secA
= tanA
...secA

Since tan A = sinA/cosA and Sec A = 1/cosA
==> tanA/SecA = (sinA/cosA) / ( cosA/SinA) =  (sinA/cosA)  * (CosA/1)

= SinA

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