Question

If sec theta +tan theta=P.Find cosec theta.

Answer 1

Secθ+tanθ=p ----------------------(1)
∵, sec²θ-tan²θ=1
or, (secθ+tanθ)(secθ-tanθ)=1
or, secθ-tanθ=1/p ----------------(2)
Adding (1) and (2) we get,
2secθ=p+1/p
or, secθ=(p²+1)/2p
∴, cosθ=1/secθ=2p/(p²+1)
∴, sinθ=√(1-cos²θ)
=√[1-{2p/(p²+1)}²]
=√[1-4p²/(p²+1)²]
=√[{(p²+1)²-4p²}/(p²+1)²]
=√[(p⁴+2p²+1-4p²)/(p²+1)²]
=√(p⁴-2p²+1)/(p²+1)
=√(p²-1)²/(p²+1)
=(p²-1)/(p²+1)
∴, cosecθ=1/sinθ=1/[(p²-1)/(p²+1)]=(p²+1)/(p²-1)

Answer 2

Answer:

$\csc\theta = \frac{P^{2}+1}{P^{2}-1}$

Step-by-step explanation:

Given:

$\sec \theta + \tan \theta = P$

We know that,

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

$\sin^{2}\theta + \cos^{2}\theta = 1$

$\sec\theta=\frac{1}{\cos\theta}$

$\tan\theta=\frac{\sin\theta}{\cos\theta}$

$\csc\theta=\frac{1}{\sin\theta}$

Now,

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

Substituting cos θ = √(1-sin²θ), we have

$P = \frac{\sqrt{1-\sin^{2}\theta}}{1-\sin\theta}$

Squaring both sides, we have

$P^{2} =\frac{(1-\sin\theta)(1+\sin\theta)}{(1-\sin\theta)^{2}}=\frac{1+\sin\theta}{1-\sin\theta}$

$P^{2} = \frac{1+\frac{1}{\csc\theta}}{1-\frac{1}{\csc\theta}} = \frac{\csc\theta+1}{\csc\theta-1}$

$P^{2}\csc\theta - P^{2} = \csc\theta + 1\\\\\csc\theta(P^{2}-1)=P^{2}+1\\\\\csc\theta = \frac{P^{2}+1}{P^{2}-1}$