Question

If sec theta + tan theta = P, obtain the values of sec theta, tan theta and sin theta in terms of P

Answer 1

hope it will be right

Answer 2

According to the question,

⇒ secθ + tanθ = p

We are asked to find the values of secθ, tanθ and sinθ in terms of p.

$\Longrightarrow \sf p = sec\theta + tan\theta$

We know that:

⇒ sec²θ = tan²θ + 1

⇒ secθ = √(tan²θ + 1)

Substitute this in p.

$\Longrightarrow \sf p = \sqrt{tan^2\theta + 1} + tan\theta$

$\Longrightarrow \sf p - tan\theta = \sqrt{tan^2\theta + 1}$

Square on both sides.

$\Longrightarrow \sf \Big[p - tan\theta\Big]^{2} = \Big[\sqrt{tan^2\theta + 1}\Big]^2$

$\Longrightarrow \sf p^2 + tan^2\theta - 2(p)(tan\theta) = tan^2\theta + 1$

$\Longrightarrow \sf p^2 + tan^2\theta - 2ptan\theta = tan^2\theta + 1$

$\Longrightarrow \sf p^2 - 2ptan\theta = 1$

$\Longrightarrow \sf p^2 - 1 = 2ptan\theta$

$\Longrightarrow \sf tan\theta = \dfrac{p^{2} - 1}{2p}$

We know that,

⇒ tanθ = opp side/adj side

⇒ Opposite side = p² - 1

⇒ Adjacent side = 2p

Using Pythogoras Theorem:

⇒ Hypotenuse² = Altitude² + Base²

⇒ Hypotenuse² = (p² - 1)² + (2p)²

⇒ H² = p⁴ + 1 - 2(p²)(1) + 4p²

⇒ H² = p⁴ + 1 - 2p² + 4p²

⇒ H² = p⁴ + 1 + 2p²

⇒ H = √[p⁴ + 1 + 2p²]

⇒ H = √[(p² + 1)²]

⇒ H = p² + 1

Now,

$\Longrightarrow \sf sec\theta = \dfrac{Hypotenuse}{Adjacent \ Side}$

$\Longrightarrow \boxed{\sf sec\theta = \dfrac{p^2 + 1}{2p}}$

$\Longrightarrow \sf tan\theta = \dfrac{Opposite \ Side}{Adjacent \ Side}$

$\Longrightarrow \boxed{\sf tan\theta = \dfrac{p^2 - 1}{2p}}$

$\Longrightarrow \sf sin\theta = \dfrac{Opposite \ Side}{Hypotenuse}$

$\Longrightarrow \boxed{\sf sin\theta = \dfrac{p^{2} - 1}{p^2 + 1}}$

Hence solved.