Question

If sec o + tan o = p show that p²-1/p²+1 = sin o​

Answer 1

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,

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

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

Hence Proved !!