Question

if
$\sec( \alpha ) + \tan( \alpha ) = p$
Then find value of
$\cosec \alpha$

Answer 1

$sec \alpha + tan \alpha = p$
$\frac{1}{cos \alpha } + \frac{sin \alpha }{ cos \alpha } = p$
$\frac{1 + \sin( \alpha ) }{ \cos( \alpha ) } = p$
$1 + \sin( \alpha ) = p \cos( \alpha )$
$\sin( \alpha ) = p \cos( \alpha ) - 1$
$\frac{1}{ \csc( \alpha ) } = p \cos( \alpha ) - 1$
$\csc( \alpha ) = \frac{1}{p \cos( \alpha ) - 1}$
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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 .