Question

in a triangle ABC , right angled at A, if tanC=√3, find the value of sinBcosC+cosBsinC

Answer 1

Answer:

$Value \: of \\ SinBCosC+cosBsinC=1$

Step-by-step explanation:

Given In ∆ABC , <A = 90°

$\angle B +\angle C = 90\degree\:---(1)$

$Value \: of \\ SinBCosC+cosBsinC\\=Sin(B+C)$

/* SinXCosY +CosXsinY =sin(X+Y) */

$=Sin 90\degree$

/* From (1) */

$= 1$

Therefore,

$Value \: of \\ SinBCosC+cosBsinC=1$

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Answer 2

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