Question

If cot²A(secA - 1)(1+cosA) = kcosA. Then find the value of k.​

Answer 1

Step-by-step explanation:

Given :-

cot²A(secA - 1)(1+cosA) = kcosA.

To find :-

Find the value of k ?

Solution :-

Given that :

cot²A(secA - 1)(1+cosA) = kcosA.

(CosA/SinA)²[(1/CosA)-1)](1+CosA)= kCos A

=>(Cos A/Sin A)² [(1-CosA)/CosA](1+CosA)

=kCosA

=>(CosA/SinA)²[(1-CosA)(1+CosA)/CosA]

= kCosA

=> (CosA/SinA)²[(1²-Cos²A)/CosA] = kCosA

Since (a+b)(a-b)=a²-b²

Where a = 1 and b = Cos A

=> (CosA/SinA)²[Sin²A/CosA] = kCosA

(Since Sin² A+ Cos² A = 1

=> Sin² A = 1 - Cos² A)

=>(Cos²A/Sin² A)(Sin² A/Cos A) = k Cos A

=>(Cos²A×Sin²A)/(Sin²A×CosA ) = k Cos A

On cancelling sin² A then

=> Cos² A / Cos A = k Cos A

=> Cos A = k Cos A

=> k Cos A = Cos A

=> k = Cos A / Cos A

=> k = 1

Therefore, k = 1

Answer:-

The value of k for the given problem is 1

Used formulae:-

• (a+b)(a-b)=a²-b²

• Sin² A+ Cos² A = 1

• Sin² A = 1 - Cos² A

• Sec A = 1 / Cos A

• Cot A = Cos A / Sin A