Question

In a ΔABC, right angled at A, if tan C = √3, find the value of sin B cos C + cos B sin C

Answer 1

Given :-

• TanC = √3

To Find :-

• sinBcosC + cosBsinC

Formula used :-

• Tan@ = (Perpendicular / Base)
• Sin@ = (Perpendicular / Hypotenuse)
• Cos@ = ( Base / Hypotenuse)

Solution :-

As, Given TanC = (√3/1) .

So, in Image we Have :-

→ Angle BAC = 90°

→ AB = √3

→ AC = 1

→ BC = Hypotenuse .

Using Pythagoras Theoram Now, we get :-

→ (Perpendicular)² + (Base)² = (Hypotenuse)²

→ (√3)² + (1)² = H²

→ H² = 3 + 1

→ H² = 4

→ H = 2.

So, Now, we Have :-

→ AB = √3, AC = 1 , BC = 2

So, Our Required Values Are :-

→ sinB = (P/H) = (AC/BC) = (1/2)

→ cosC = (B/H) = (AC/BC) = (1/2)

→ cosB = (B/H) = (AB/BC) = (√3/2)

→ sinC = (P/H) = (AB/BC) = (√3/2)

Putting All These Value in Our Question Now :-

→ sinBcosC + cosBsinC

→ (1/2) * (1/2) + (√3/2) * (√3/2)

→ (1/2)² + (√3/2)²

→ (1/4) + (3/4)

→ (4/4)

→ 1 (Ans).

Answer 2

$\large \underline{ \green{ \boxed{ \bf \blue{Required \: Answer}}}}$