Question

if secA + tanA = p then find the value of secA

Answer 1

secA+tanA=p ----------- (i)

‘.' sec²A-tan²A=1 (secA+tanA)(secA-tanA)=1

secA-tanA = 1/p ------------- (ii)

Subtracting eqn ii from I
We get,
2tanA= p-1/p
tanA = (p²-1)/2p
cotA = 2p/(p²-1)
Now
cosec²A-cot²A = 1
cosec²A=1+cot²A
cosec²A=1+[2p/(p²-1)]²
cosec²A= 1+4p²/(p²-1)²
cosec²A= (p²)² - 2p²- 1- 4p²/(p²-1)²
cosec²A= (p²)² + 2p²+1/(p²-1)²
cosec²A= (p²+1)² / (p²-1)²
cosecA= (p²+1)/p²-1)

Answer 2

Answer:

Given :

sec A + tan A = p

I am replacing p by ' k '

sec A + tan A = k

We know :

sec A = H / B   & tan A = P / B

H / B + P / B =  k / 1

H + P / B =  k / 1

So , B = 1

H + P = k

P = k - H

From pythagoras theorem :

H² = P² + B²

H² = ( H - k )² + 1

H² = H² + k² - 2 H k + 1

2 H k = k² + 1

H = k² + 1 / 2 k

P = k - H

P = k² - 1 / 2 k

Now write k = p we have :

Base = 1

Perpendicular P = P² - 1 / 2 P

Hypotenuse H = P² + 1 / 2 P

Value of sec A = H / B

sec A =  P² + 1 / 2 P / 1

sec A =  P² + 1 / 2 P

Therefore , we got value .