Question

If seca +tana=p then find the value of coseca

Answer 1

$\sec(a) + \tan(a) = p$

$\frac{1}{ \cos(a) } + \frac{ \sin(a) }{ \cos(a) } = p$

1+sina/cosa=p

1+sina=pcosa

1+sina=p(root)(1- sin^2a)

squaring on both side

(1+sina)^2=p^2 (1-sin^2a)

1+2sina+sin^2a=p^2-p^2sin^2a

1+2sina+sin^2a-p^2+p^2sin^2a=0

sin^2 (1+p^2)+2sina+(1-p^2)=0

a=(1+p^2) b=2 c=(1-p^2)

D=b^2-4ac

D= (2)^2-4×(1+p^2)×(1-p^2)

= 4-4(1)^2-(p^2)^2

=4-4+4p^2

D= 4p^2

By quadratic formula,

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 .