In a ∆ABC, right angled at A, if tan C = √3 , find the value of sin B cos C + cos B sin C.
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Answer:
The value of sinBcosC+cosBsinC is 1
Step-by-step explanation:
Step 1:
Given Data:
To Find the Value of sinBcosC+cosBsinC if tanC=√3:
∠A=90 °(given)
∠A+ ∠B+ ∠C=180° (sum of the interior angle of the triangle)
Step 2:
90+ ∠B+ ∠C=180°
∠B+∠C=90°
Step 3:
∠B=90°- ∠C
So,
Step 4:
sinBcosC+cosBsinC
Step 5:
=sin(90-C)cosC+cos(90-C)sinC
=cosC x cosC+sinC+sinC
Therefore [sin(90- θ) =cos θ]
[cos(90- θ) =sin θ]
Step 6:
=Cos2C+sin2C
=Sin2C+cos2C (Sin2 θ +cos2 θ)
=1
Step 7:
The value of sinBcosC+cosBsinC is 1
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