Question

If secA+tanA=p then find cosecA

Answer 1

hey mate here is your answer ✌♥✌♥

secA+tanA=p.......1

now as we know that

sec²A-tan²A=1

so by using a²-b²=(a+b)(a-b)

(secA+tanA)(secA-tanA)=1

p(secA-tanA)=1

(secA-tanA)=1/p.......2

now......


Subtracting (2) from (1) we get,

2tanA=p-1/p

or, tanA=(p²-1)/2p

∴, cotA=2p/(p²-1)

Now, cosec²A-cot²A=1

or, cosec²A=1+cot²A

or, cosec²A=1+{2p/(p²-1)}²

or, cosec²A=1+4p²/(p²-1)²

or, cosec²A=(p⁴-2p²+1+4p²)/(p²-1)²

or, cosec²A=(p⁴+2p²+1)/(p²-1)²

or, cosec²A=(p²+1)²/(p²-1)²

or, cosecA=(p²+1)/(p²-1)

hope it will be helpful to you ✔✔✔✔

mark me brainliest ✌✌✌

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 cosec A = H / P

cosec A =  P² + 1 / 2 P / P² - 1 / 2 P

cosec A = P² + 1 / P² - 1

Therefore , we got value .