Question

Guys please tell this

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Answer 1

Answer:

1

Step-by-step explanation:

Solving in parts,

Solving cos(40° - A) - sin(50° + A):

Using property, cosA = sin(90 - A)

cos(40° - ∅) = sin(90 - {40 - A})

= sin(90 - 40 + A)

= sin(50 + A) ...(1)

Thus, cos(40° - A) - sin(50° + A)

= sin(50° + A) - sin(50° + A) {(1)}

= 0

Solving (cos²40 + cos²50):

As mentioned earlier, cosA = sin(90-A)

=> cos²(40) + sin²(90 - 50)

=> cos²(40) + sin²(40)

=> 1 {sin²A + cos²A = 1}

I changed cos²50 only, since it is clearly visible that later on sin²A + cos²A = 1 can be applied.

Similarly with the denominator: we get 1.

Now,

=> 0 + 1/1

=> 1

Answer 2

Answer:

1

Step-by-step explanation:

Solving in parts,

Solving cos(40° - A) - sin(50° + A):

Using property, cosA = sin(90 - A)

cos(40° - ∅) = sin(90 - {40 - A})

= sin(90 - 40 + A)

= sin(50 + A) ...(1)

Thus, cos(40° - A) - sin(50° + A)

= sin(50° + A) - sin(50° + A) {(1)}

= 0

Solving (cos²40 + cos²50):

As mentioned earlier, cosA = sin(90-A)

=> cos²(40) + sin²(90 - 50)

=> cos²(40) + sin²(40)

=> 1 {sin²A + cos²A = 1}

I changed cos²50 only, since it is clearly visible that later on sin²A + cos²A = 1 can be applied.

Similarly with the denominator: we get 1.

Now,

=> 0 + 1/1

=> 1

Hope this answer will help you.✌️