Question

If secA +tanA= p then prove that sinA =(p2 - 1)/(p2+ 1)

Answer 1

secA+tanA=p ----------------------------(1)
We know that,
sec²A-tan²A=1
or, (secA+tanA)(secA-tanA)=1
or, p(secA-tanA)=1
or, secA-tanA=1/p -----------------------(2)
Adding (1) and (2) we get,
2secA=p+1/p
or, secA=(p²+1)/2p
∴, cosA=1/secA=2p/(p²+1)
∴, sinA=√(1-cos²A)
=√{1-4p²/(p²+1)²
=√{(p²+1)²-4p²}/(p²+1)²
=√(p⁴+2p²+1-4p²)/(p²+1)
=√(p²-1)²/(p²+1)
=(p²-1)/(p²+1) (Proved)

Answer 2

Answer:

We are given that $secA+tanA=p$  ----A

We know that $sec^2A-tan^2A=1$

Using identity ; $(a+b)(a-b)=a^2-b^2$

$(secA+tanA)(secA-tanA)=1$

Using A

$p(secA-tanA)=1$

$(secA-tanA)=\frac{1}{p}$ ---- B

Adding A and B we get,

$2secA=p+\frac{1}{p}$

$secA=\frac{p^2+1}{2p}$

$cosA=\frac{1}{secA}=\frac{2p}{(p^2+1)}$

We now that $Sin^2 A + Cos^2 A = 1$

$sinA=\sqrt{(1-cos^2A)}$

$\sqrt{1-\frac{4p^2}{(p^2+1)^2}}$

$\sqrt{\frac{(p^2+1)^2-4p^2}{(p^2+1)^2}}$

$\frac{\sqrt{p^4+2p^2+1-4p^2}}{p^2+1}$

$\frac{\sqrt{(p^2-1)^2}}{(p^2+1)}$

$\frac{(p^2-1)}{(p^2+1)}$

Hence proved