Question

secA+tanA=p
then prove that
p^2-1/p^2+1= sinA​

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:

secA+tanA=p .....(i)

We know sec2A–tan2A=1

=>(secA–tanA)(secA+tanA)=1

=>(secA–tanA)(p)=1

=>secA–tanA=1p .....(ii)

Adding (i) and (ii)

secA+tanA+secA–tanA=p+1p

=>2secA=p2+1p

=>secA=p2+12p

=>cosA=2pp2+1

=>1–sin2A−−−−−−−√=2pp2+1

=>sin2A=1−(2pp2+1)2

=>sin2A=1−(4p2(p2+1)2)

=>sin2A=((p2+1)2−4p2(p2+1)2)

=>sin2A=((p2−1)2(p2+1)2)

=>sin2A=(p2−1p2+1)2

=>sinA=p2−1p2+1

=>cosecA=p2+1p2−1