Question

If sec teta + tan teta=p find the value of cosec teta

Answer 1

$\huge\mathfrak\purple{Solution}$

sec∅ + tan∅ = p

Find out

cosec∅ = ?

We know that

$\underline{\underline{According \: to \: trigonometry \: identies } }$

$\boxed{\boxed{ {sec}^{2}\theta - {tan}^{2}\theta = 1} }$

Then,

$\sf\implies\: {sec}^{2}\theta - {tan}^{2}\theta = 1 \\ \\ \sf\implies \: (sec\theta - tan\theta )(sec\theta + tan\theta ) = 1 \\ \\ \sf\implies \: (sec\theta - tan\theta ) = \frac{1}{p}$

So,

$sec\theta = \frac{1}{2}(p + \frac{1}{p} ) \\ \\ sec\theta= \frac{{p}^{2} + 1}{2p} \\ \\ tan\theta = \frac{1}{2}(p - \frac{1}{p}) \\ \\ tan\theta = \frac{{p}^{2} - 1}{2p}$

Therefore,

$cosec\theta = \frac{1}{sin\theta} \\ \\ = \frac{\frac{1}{cos\theta}}{\frac{sin\theta}{cos\theta}} \\ \\ = \frac{sec\theta}{tan\theta} \\ \\ <strong>\</strong><strong>:</strong><strong>putt</strong><strong>ing</strong><strong>\</strong><strong>:</strong><strong> </strong><strong>the</strong><strong> </strong><strong>\</strong><strong>:</strong><strong> </strong><strong>value</strong> \\ \\ = \frac{\frac{{p}^{2} +1}{2p}}{\frac{{p}^{2} - 1}{2p}} \\ \\ cosec\theta = \frac{{p}^{2} + 1}{{p}^{2} - 1 }$

Hence,

$\huge\boxed{value \:of \:cosec\theta = \frac{{p}^{2} + 1}{{p}^{2} - 1} }$

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 cosec A = H / P

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

cosec A = P² + 1 / P² - 1

Therefore , we got value .