Question

solve the given problem

Answer 1

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



$\underline{\textsf{Given,}} \\ \\ \sf \implies cosec \: \theta \: + \: cot \: \theta \: = \: P \qquad...(1)$




$\textsf{Using Trigonometry Identity : }\\ \\ \sf \implies cosec^2 \theta \: - \: cot^2 \theta \: = \: 1 \\ \\ \sf \implies (cosec \: \theta \: + \: cot \: \theta)(cosec \: \theta \: - \: cot \: \theta ) \: = \: 1 \\ \\ \textsf{Plug the value of (1),}$




$\sf \implies P( cosec \: \theta \: - \: cot \: \theta) \: = \: 1 \\ \\ \sf \: \: \therefore \: \: cosec \: \theta \: - \: cot \: \theta \: = \: \dfrac{1}{P} \quad...(2)$




$\textsf{Add (1) and (2), } \\ \\ \sf \implies cosec \: \theta \: + \: cot \: \theta \: + \: cosec \: \theta \: - \: cot \: \theta \: = \: P \: + \: \dfrac{1}{P} \\ \\ \sf \implies 2 \: cosec \: \theta \: = \: \dfrac{P^2 \: + \: 1}{P} \\ \\ \sf \: \: \therefore \: \: cosec \: \theta \: = \: \dfrac{P^2 \: + \: 1}{2P}$




$\textsf{Subtract (2) from (1), } \\ \\ \sf \implies (cosec \: \theta \: + \: cot \: \theta )\: - \: (cosec \: \theta \: - \: cot \: \theta )\: = \: P \: - \: \dfrac{1}{P} \\ \\ \sf \implies cosec \: \theta \: + \: cot \: \theta \: - \: cosec \: \theta \: + \: cot \: \theta \: = \: \dfrac{P^2 \: - \: 1}{P} \\ \\ \sf \implies 2 \: cot \: \theta \: = \: \dfrac{P^2 \: - \: 1}{P} \\ \\ \sf \: \: \therefore \: \: cot \: \theta \: = \: \dfrac{P^2 \: - \: 1}{2P}$




$\underline{\textsf{Now,}} \\ \\ \sf \implies cos \: \theta \: = \: cot \: \theta \: \div \: cosec \: \theta \\ \\ \sf = \dfrac{P^2 \: - \: 1}{2P} \: \div \: \dfrac{P^2 \: + \: 1 }{2P} \\ \\ \sf = \dfrac{P^2 \: - \: 1}{2P} \: \times \: \dfrac{2P}{P^2 \: + \: 1 } \\ \\ \sf = \dfrac{P^2 \: - \: 1}{P^2 \: + \: 1 }$




$\textsf{Thanks to @Happiestwriter012 for this easy method !!}$

Answer 2

Cosec A + Cot A = P

= 1/Sin A + Cos A / Sin A = P

= 1+Cos A/Sin A = p

=> SQUARING ON BOTH SIDES

= (1+COS A)²/(SINA)²= P²

= (1+ Cos A)² / (Sin A)² = P²

= (1+cos A)² = (p²)[(Sin A)²]

= (1+ cos A) ² = (p²) [(1-cos²A)]

= (1+ cos A) ² = (p²) [ (1+cosA)(1-cos A) ]

= (1+cos A)² ÷ (1+ cos A) = (p²)[(1-cos A)]

= 1+cos A = (p²)[1-cos A]

= 1+cos A ÷ 1-cos A = p²

= Here, Using (a+b/a-b=c+d/c-d). This is known as Componendo and dividendo

According to the Question statement!


1+cos A ÷ 1-cos A = p²

Then,

(1+cos A) + (1 - cos A ) ÷ (1+ Sin A - (1-sin A) = p²+1 / p²-1

= 2/2cos = p²+1/p²-1

= 1/cos = p²+1/p²-1

= Sec = p²+1 /p²-1

We know that Cos A = 1/sec A

Then,

Cos A = p²-1 /p²+1