Question

if secA+tanA=p, find:p2-1/p2+1

Answer 1

$\bf\large\underline\pink{Solution:-}$

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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)

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Answer 2

$\huge\purple{Answer:-}$

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)