Question

SinA+CosA=√2
TanA+CotA=2
Then find,
1) SinA×CosA
2) SecA×CosecA
3) Sin^2A-Cos^2A

Answer 1

Answer:-

Given :

Sin A + Cos A = √2

(Square on both sides)

→ (Sin A + Cos A)² = (√2)²

Using (a + b)² = a² + b² + 2ab

→ Sin² A + Cos² A + 2*Sin A * Cos A = 2

By using Sin² A + Cos² A = 1

→ 1 + 2 * Sin A * Cos A = 2

→ 2*Sin A * Cos A = 2 - 1

→ Sin A * Cos A = 1/2 (1)

→ 1/Cosec A * 1/Sec A = 1/2

After cross multiplication we get,

→ 2 = Sec A * Cosec A

→ Sec A * Cosec A = 2 (2)

We know that,

a² - b² = (a + b)(a - b)

To find (a - b) i.e., Sin A - Cos A

We know,

(a - b)² = (a + b)² - 4ab

→ (Sin A - Cos A)² = (Sin A + Cos A)² - 4*Sin A* Cos A.

→ (Sin A - Cos A)² = (√2)² - 4(1/2)

→ (Sin A - Cos A)² = 2 - 2 = 0

→ Sin A - Cos A = 0

Now,

Sin² A - Cos² A = (Sin A + Cos A)(Sin A - Cos A)

→ Sin² A - Cos² A = (√2)(0)

→ Sin² A - Cos² A = 0 (3)