Question

if sect theta + tan theta =p
then find value of sec theta-tan theta​

Answer 1

Answer:

Given : secx + tanx = p ................................... (i)

Now, we know that : sec²x - tan²x = 1

or, (secx + tanx)(secx - tanx) = 1

Putting (i) in the equation, we get :-

p(secx - tanx) = 1

or, secx - tanx = 1/p .........................................(ii)

Now adding (i) and (ii), we get :-

secx + tanx + secx - tanx = 1/p + p

or, 2secx = (1+p²)/p

or, secx = (1+p²)/2p   

On subtracting (ii) from (i), we get :-

(secx + tanx) - (secx - tanx) = p - 1/p

or, 2tanx = (p² - 1)/p

or, tanx = (p² - 1)/2p    

We know that sinx = sinx/cosx x cosx

or, sinx = tanx x 1/secx

or, sinx = tanx/secx

or, sinx = [(p² - 1)/2p]/[(p² + 1)/2p] ..(In the next step both the 2p get cancelled)

or, sinx = (p² - 1)/(p² + 1)  

cosecx = 1/sinx

= 1/ [(p² - 1)/(p² + 1)]

= (p²+1)/(p²-1)

Answer 2

$\huge{\underline{\underline{\green{\mathsf{Answer:}}}}}$

$\huge\fbox{\mathbf{\frac{1}{p}}}$

$\huge{\underline{\underline{\green{\mathsf{Explaination:}}}}}$

Given

$\mathsf{sec\theta + tan\theta = p}$

We know that

$\mathsf{sec^2\theta = 1 + tan^2\theta}$

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

$\fbox{\mathsf{Using\: :a^2 - b^2 = (a+b) (a-b)}}$

$\mathsf{\implies (sec\theta + tan\theta)(sec\theta - tan\theta) = 1}$

$\mathsf{\implies \: p (sec\theta - tan\theta) = 1}$

$\mathsf{\implies \: sec\theta - tan\theta = \frac{1}{p}}$