Question

if sec a + tan a = p then show that p^2 -1/p^2+1 = sin a​

Answer 1

Answer:  SecA +TanA = p------(1)

⇒Sec²A - Tan²A = 1 (trignometry identity)

⇒(SecA + TanA)(SecA - TanA) = 1

⇒SecA - TanA = 1/(SecA + TanA) = 1/p----------(2)

⇒equation(1+2)  secA + tanA + secA - TanA= p +1/p

                            2secA = p + 1/p-----------(3)

⇒equation(1-2)  secA + tanA - secA + TanA= p -1/p

                            2TanA = p - 1/p-----------(4)

∴$\frac{eq4}{eq3}=$$\frac{tanA}{secA}$

dividing equation 3 by equation 4

= 2TanA/2SecA= (p-1/p)/(p+1/p)

= SinA = $\frac{p^2-1}{p}$/$\frac{p^2+1}{p}$

= sinA = p²-1/p²+1

hence proved

Step-by-step explanation: