Question

(b) If sec A+ tan A = p, show that: sin A
p^2-1
p²+1​

Answer 1

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

Given:-

Sec A + Tan A = p

To find:-

Show that Sin A = (p^2-1)/(p^2+1)

Solution:-

Given that :-

Sec A + Tan A = P --------(1)

We know that

Sec^2 A - Tan^2 A = 1

=>(Sec A + Tan A )(Sec A - Tan A )=1

=>P(Sec A - Tan A) = 1

Sec A - Tan A = 1/P -------(2)

Now , On adding (1)&(2)

Sec A + Tan A = P

Sec A - Tan A = 1/P

(+)

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2 Sec A + 0 = P+(1/P)

_________________

2 Sec A = P+(1/P) -------------(3)

On Subtracting (2) from (1)

Sec A + Tan A = P

Sec A - Tan A = 1/P

(-)

________________

0+2 Tan A = P-(1/P)

________________

2 Tan A = P-(1/P) -----------(4)

On dividing (4) by (3)

2 Tan A/2 Sec A = [P-(1/P)]/[P+(1/P)]

=>Tan A/ Sec A= [(P^2-1)/P]/[(P^2+1)/P]

=>Tan A / Sec A = (P^2-1)/(P^2+1)

=>(Sin A/Cos A)/(1/ Cos A) = (P^2-1)/(P^2+1)

=> Sin A = (P^2-1)/(P^2+1)

Hence, Proved

Answer:-

Sin A = (P^2-1)/(P^2+1) for the given problem

Used formulae:-

• Tan A = Sin A/ Cos A
• Sec A = 1/ Cos A
• Sec^2 A - Tan^2 A = 1