Question

sec + tan = p find value of cosec​

Answer 1

Solution :-

secθ + tanθ = p----eq(1)

We know that

sec²θ - tan²θ = 1

⇒ (secθ + tanθ)(secθ - tanθ) = 1

[ Beause x² - y² = (x + y)(x - y) ]

⇒ p(secθ - tanθ) = 1

⇒ secθ - tanθ = 1/p ----eq(2)

Adding eq(1) and eq(2)

⇒ secθ + tanθ + (secθ - tanθ) = p + 1/p

⇒ secθ + tanθ + secθ - tanθ = (p² + 1)/p

⇒ 2secθ = (p² + 1)/p

⇒ secθ = (p² + 1)/2p

Subracting eq(2) from eq(1)

⇒ secθ + tanθ - (secθ - tanθ) = p - 1/p

⇒ secθ + tanθ - secθ + tanθ = (p² - 1)/p

⇒ 2tanθ = (p² - 1)/p

⇒ tanθ = (p² - 1)/2p ---eq(4)

Dividing eq(3) by eq(4)

⇒ secθ/tanθ = [ (p² + 1)/2p ] / [ (p² - 1)/2p ]

⇒ [ 1/cosθ ] / [ sinθ/cosθ ] = (p² + 1) / (p² - 1)

⇒ 1/sinθ = (p² + 1) / (p² - 1)

⇒ cosecθ = (p² + 1) / (p² - 1)

Hence, the value of cosecθ is (p² + 1) / (p² - 1).

Answer 2

$\Large\underline\mathfrak{Question}$

• if sec A + tan A = p
• Find cosec A = ?

$\Large\bold\star\underline{\underline\textbf{Formula\:used}}$

• (a-b)² = a² + b² + ab
• sec²A - tan²A = 1
• Hypotenuse² = perpendicular ² + Base²
• Tan @ = P/B
• cosec A = H/P

$\Large\underline{\underline{\sf{Solution}:}}$

__________________

$\red{\textbf{My Approach in Easiest way}} \:$

$secA + tanA = p \\ \\ \red\leadsto \: secA = (p - tanA) \\ \\ squaring \: both \: sides \\ \\ \red\leadsto \: sec^{2} A = {p}^{2} - 2ptanA + tan^{2} A \\ \\ \red\leadsto \: sec^{2} A - tan^{2} A = {p}^{2} - 2ptanA \\ \\ \red\leadsto \: 1 = {p}^{2} - 2ptanA \\ \\ \red\leadsto \: \large\boxed{\bold{tanA = \frac{ {p}^{2} - 1}{2p} }}$

Now, we know that,

→ Tan A = P/B

Hence,

→ Perpendicular = (P²-1)

→ Base = 2P

so, By pythagoras theoram we get,

$H^{2} = (p^{2} -1)^{2} + (2p)^{2} \\ \\ \red\leadsto \: H^{2} = \: {p}^{4} + 1 - 2 {p}^{2} + 4 {p}^{2} \\ \\\red\leadsto \: H^{2} = \: {p}^{4} + 1 + 2 {p}^{2} \\ \\ \red\leadsto \: H^{2} = ( {p}^{2} + 1) ^{2} \\ \\ \green{\large\boxed{\bold{H = ( {p}^{2} + 1)}}}$

Now, we know that,

$CosecA = \frac{H}{P} \\ \\ hence \\ \\\pink{\large\boxed{\boxed{\bold{ CosecA \: = \frac{ {p}^{2} + 1}{ {p}^{2} - 1} }}}}$

$\large\underline\textbf{Hope it Helps You.}$