Question

if (1+tanα tanβ)^2+(tanα-tanβ)^2=k,then 2k.cos^2α.cos^2βis​

Answer 1

Step-by-step explanation:

Given :-

(1+Tan α Tan β)²+(Tan α-Tan β)²=k

To find :-

Find the value of 2k.Cos²α . Cos²β ?

Solution :-

Given that

(1+Tan α Tan β)²+(Tan α-Tan β)²=k

Now,

On taking (1+Tan α Tan β)²

=> 1² + Tan² α Tan² β + 2 Tan α Tan β

Since (a+b)² = a²+2ab+b²

=> 1+ Tan² α Tan² β + 2 Tan α Tan β ------(1)

On taking (Tan α-Tan β)²

=> Tan² α+Tan² β -2 Tan α Tan β -------(2)

Since (a-b)² = a²-2ab+b²

On adding (1) & (2) then

1+ Tan² α Tan² β + 2 Tan α Tan β + Tan² α+Tan² β -2 Tan α Tan β

=> 1+ Tan² α Tan² β + Tan² α+Tan² β

=> (1+Tan² α )+Tan² β ( Tan² α+1)

=> (1+Tan² α )(1+Tan² β )

=> Sec² α Sec² β

Since Sec² A - Tan² A = 1

Now,

(1+Tan α Tan β)²+(Tan α-Tan β)²=k

=> Sec² α Sec² β = k

=> (1/Cos²α)(1/ Cos² β ) = k

=> 1 / (Cos²α Cos² β ) = k

=> 1 = k (Cos²α Cos² β )

On multiplying with 2 both sides then

=> 2 = 2k Cos²α Cos² β

=> 2k Cos²α Cos² β = 2

Answer:-

The value of 2k Cos²α Cos² β for the given problem is 2

Used formulae:-

→ (a+b)² = a²+2ab+b²

→ Sec² A - Tan² A = 1

→ Sec A = 1/ Cos A

→(a-b)² = a²-2ab+b²