Question

If secA +tanA=P , show 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)

hope this helps :-)

Answer 2

hi friend, I'll solve the sum and post it.... hope this helps you and beat of luck for math.