Question

If sec theta + tan theta = p , prove that sin theta = p 2 - 1 / p 2 + 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:

Sec A + Tan A = p

To Prove: Sin A = ( p² - 1 / p² + 1 )

Proof:

Calculating p² - 1 we get,

⇒ ( Sec A + Tan A )² - 1

⇒ ( Sec²A + Tan²A + 2 SecA.TanA - 1 )

⇒ ( Sec²A - 1 + Tan²A + 2 SecA.TanA )

We know the identity that, Sec²A - 1 = Tan²A. Substituting that we get,

⇒ ( Tan²A + Tan²A + 2SecA.TanA )

⇒ ( 2 Tan²A + 2 SecA.TanA )

Taking 2 TanA common we get,

⇒ 2 TanA ( Tan A + Sec A )   ... ( Equation 1 )

Now solving the denominator ( p² + 1 ) we get,

⇒ ( Sec A + Tan A )² + 1

⇒ ( Sec²A + Tan²A + 2SecA.TanA + 1 )

We know that, 1 + Tan²A = Sec²A. Hence substituting that we get,

⇒ ( 1 + Tan²A + Sec²A + 2 SecA.TanA )

⇒ ( Sec²A + Sec²A + 2 SecA.TanA )

⇒ ( 2 Sec²A + 2 SecA.TanA )

Taking 2 SecA common we get,

⇒ 2 SecA ( Sec A + Tan A ) .... ( Equation 2 )

Now dividing Equation 1 by Equation 2 we get,

$\implies \dfrac{ 2\: TanA\: ( Tan A + Sec A ) }{ 2\: SecA \:( Sec A + Tan A ) }$

( Tan A + Sec A ) will get cancelled and we get,

$\implies \dfrac{ 2\: Tan A }{2 \: Sec A} \implies \dfrac{Tan A}{SecA}$

Writing both of them in terms of SinA and CosA we get,

$\implies \dfrac{\dfrac{SinA}{CosA}}{ \dfrac{1}{CosA} }\\\\\\\\\implies \dfrac{SinA}{CosA} \times \dfrac{CosA}{1} \\\\\\\text{CosA gets cancelled and we get,}\\\\\implies \dfrac{SinA}{1} = SinA$

Hence Proved !!