Question

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Answer 1

Question

If secθ + tanθ = p, then find the value of cosecθ.

Solution

$sec \theta + tan \theta = p \longrightarrow (1)$

We know that

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

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

[ Because a² - b² = (a + b)(a - b) ]

$\implies p \times (sec \theta - tan \theta)=1$

$\implies sec \theta - tan \theta= \dfrac{1}{p} \longrightarrow(2)$

Adding (1) & (2)

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

$\implies sec \theta + tan \theta + sec \theta - tan \theta = \dfrac{ {p}^{2} + 1 }{p}$

$\implies 2sec \theta = \dfrac{ {p}^{2} + 1 }{p}$

$\implies sec \theta = \dfrac{ {p}^{2} + 1 }{2p} \longrightarrow(3)$

Subtracting (2) from (1)

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

$\implies sec \theta + tan \theta - sec \theta + tan \theta= \dfrac{ {p}^{2} - 1}{p}$

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

$\implies tan \theta= \dfrac{ {p}^{2} - 1}{2p} \longrightarrow(4)$

Dividing (3) by (4)

$\implies sec \theta \div tan \theta = \dfrac{ {p}^{2} + 1 }{2p} \div \dfrac{ {p}^{2} - 1 }{2p}$

$\implies \dfrac{1}{cos \theta} \div \dfrac{sin \theta}{cos \theta} = \dfrac{ {p}^{2} + 1 }{2p} \times \dfrac{2p}{ {p}^{2} - 1}$

[ Because secθ = 1/cosθ and tanθ = sinθ/cosθ ]

$\implies \dfrac{1}{cos \theta} \times \dfrac{cos \theta}{sin \theta} = \dfrac{ {p}^{2} + 1 }{ {p}^{2} - 1 }$

$\implies \dfrac{1}{sin \theta} = \dfrac{ {p}^{2} + 1 }{ {p}^{2} - 1 }$

$\implies cosec \theta = \boxed{\dfrac{ {p}^{2} + 1 }{ {p}^{2} - 1 } }$

[ Because 1/sinθ = cosecθ ]

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

Answer 2

Given :---

• secA + tanA = p

To Find :---

• cosecA ?

Formula used :---

• sec²A - tan²A = 1
• cotA = 1/tanA
• cosec²A - cot²A = 1
• (a² - b²) = (a+b)(a-b)
• (a²+2ab+b²) = (a+b)²

Solution :-------

secA+tanA = p ---------------------------- Equation (1)

using formula here, sec²A-tan²A = 1

and (a² - b²) = (a+b)(a-b)

→ (secA+tanA)(secA-tanA) = 1

→ secA-tanA = 1/p -----------------------Equation (2)

Subtracting Equation (2) from Equation (1) we get,

2tanA = p-(1/p)

→ tanA = (p²-1)/2p

using cotA = 1/tanA

→ cotA = 2p/(p²-1) ---------------------- Equation (3)

Now, using cosec²A-cot²A = 1

→ cosec²A = 1+cot²A

Putting value of cotA from Equation (3)

→ cosec²A = 1+[2p/(p²-1)]²

→ cosec²A = 1+4p²/(p²-1)²

→ cosec²A = (p⁴-2p²+1+4p²)/(p²-1)²

→ cosec²A = (p⁴+2p²+1)/(p²-1)²

→ cosec²A = (p²+1)²/(p²-1)²

→ cosecA = (p²+1)/(p²-1) (Ans)

(Hope it Helps you)