Question

sec theta -tan theta=p+1/p-1 then find cos theta

Answer 1

Answer:

Step-by-step explanation:

secθ-tanθ=p+1/p-1----------------(1)

sec²θ-tan²θ=1

sec²θ-tan²θ/secθ-tanθ=1/p+1/p-1=p-1/p+1

secθ+tanθ=p-1/p+1----------------(2)

adding (1)  and (2)

2secθ=p+1/p-1-+p-1/p+1

=1/p²-1(p+1+p-1)

2secθ=2p/p²-1

secθ=p/p²-1

cosθ=p²-1/p

Answer 2

$\mathfrak{\underline{\underline{\green{Answer:-}}}}$

$cosA = \dfrac{{p}^{2}-1}{{p}^{2}+1}$

$\mathfrak{\underline{\underline{\green{Explanation:-}}}}$

Given:-

$\mathsf{ secA-tanA = \dfrac{p+1}{p-1}}$

To Find:-

$\mathsf{cosA }$

Solution:-

$\mathsf{Let, \: secA-tanA = \dfrac{p+1}{p-1} ------(1)}$

As we know,

$\star\boxed{\red{{sec}^{2}A-{tan}^{2}A= 1}}$

So,

$\mathsf{ {sec}^{2}A-{tan}^{2}A = 1}$

By the Identity

$\star\boxed{\red{{a}^{2}-{b}^{2}=(a+b)(a-b) }}$

$\mathsf{(secA+tanA)(secA-tanA) = 1}$

$\mathsf{(secA+tanA)(\dfrac{p+1}{p-1} ) = 1}$

$\mathsf{ secA+tanA=\dfrac{p-1}{p+1} --------(2)}$

Now, add (1) and (2)

$\mathsf{ secA-tanA +secA+tanA = \dfrac{p+1}{p-1} + \dfrac{p-1}{p+1}}$

$\mathsf{ 2secA = \dfrac{{(p+1)}^{2} + {(p-1)}^{2} }{(p+1)(p-1)} }$

$\mathsf{ 2secA = \dfrac{{p}^{2}+1+2p+{p}^{2}+1-2p}{(p+1)(p-1)} }$

$\mathsf{ 2secA = \dfrac{2{p}^{2}+2}{{p}^{2} -1}}$

$\mathsf{ secA = \dfrac{2({p}^{2}+1)}{2({p}^{2}-1)} }$

$\mathsf{ secA = \dfrac{\cancel{2}({p}^{2}+1)}{\cancel{2}({p}^{2}-1)} }$

$\mathsf{ secA = \dfrac{{p}^{2}+1}{{p}^{2}-1} }$

$\star \boxed{\red{secA= \dfrac{1}{cosA}}}$

$\mathsf{ \dfrac{1}{cosA} = \dfrac{{p}^{2}+1}{{p}^{2}-1} }$

$\mathsf{ cosA = \dfrac{{p}^{2}-1}{{p}^{2}+1} }$

Hence,

$\mathsf{ cosA = \dfrac{{p}^{2}-1}{{p}^{2}+1} }$

$\mathbb{\pink{NOTE:-}}$

Assume 'A' as 'Theta'