If A + B = 45° then find the value of sinASinB + (1 + tanA)(1 + tanB)

1.	If A + B = 45° then find the value of sinASinB + (1 + tanA)(1 + tanB) – cosAcosB1. (2√2 – 1)/22. (2√2 + 1)/23. (2√2 – 1)/√24. (2√2 + 1)/√2
Answer» Correct Answer - Option 3 : (2√2 – 1)/√2 Given: A + B = 45° Concept used: 1.) cos(A + B) = cosAcosB – sinAsinB 2.) If A + B = 45° then (1 + tanA)(1 + tanB) Calculations: sinASinB + (1 + tanA)(1 + tanB) – cosAcosB ⇒ (1 + tanA)(1 + tanB) – cosAcosB + sinAsinB ⇒ (1 + tanA)(1 + tanB) – (cosAcosB – sinAsinB) ⇒ (1 + tanA)(1 + tanB) – cos(A + B) Here A + B = 45° ⇒ 2 – cos(45°) ⇒ 2 – 1/√2 ⇒ (2√2 – 1)/√2 ∴ The correct answer is (2√2 – 1)/√2

If A + B = 45° then find the value of sinASinB + (1 + tanA)(1 + tanB) – cosAcosB1. (2√2 – 1)/22. (2√2 + 1)/23. (2√2 – 1)/√24. (2√2 + 1)/√2

Answer» Correct Answer - Option 3 : (2√2 – 1)/√2

Given:

A + B = 45°

Concept used:

1.) cos(A + B) = cosAcosB – sinAsinB

2.) If A + B = 45° then (1 + tanA)(1 + tanB)

Calculations:

sinASinB + (1 + tanA)(1 + tanB) – cosAcosB

⇒ (1 + tanA)(1 + tanB) – cosAcosB + sinAsinB

⇒ (1 + tanA)(1 + tanB) – (cosAcosB – sinAsinB)

⇒ (1 + tanA)(1 + tanB) – cos(A + B)

Here A + B = 45°

⇒ 2 – cos(45°)

⇒ 2 – 1/√2

⇒ (2√2 – 1)/√2

∴ The correct answer is (2√2 – 1)/√2