Question

secA + tanA = p , find the value of p

Answer 1

i think question data is not sufficient to solve this question
secA+tanA=p......eq(1),

since
sec²A-tan²A=1,

then
(secA+tanA).(secA-tanA)1

p.(secA-tanA)=1,

then
(secA-tanA)=1/p,......eq(2),

adding equation 1st and 2nd we get,

2secA=p+1/p

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 .