Question

IF Tan A =1/√3 and Tan B=√3, then find Sin A.Cos B+Cos A.sin B.. ?

Answer 1

Given:

↬ $\bf{tanA=\dfrac{1}{\sqrt{3}}}$

↬ $\bf{tanB=\sqrt{3}}$

To Find:

• Value of sinA.cosB+cosA.sinB

Solution:

We know that,

$\longrightarrow\bf{tan30^{\circ}=\dfrac{1}{\sqrt{3}}}$

$\longrightarrow\bf{tan60^{\circ}=\sqrt{3}}$

$\longrightarrow\bf{sin30^{\circ}=\dfrac{1}{2}}$

$\longrightarrow\bf{sin60^{\circ}=\dfrac{\sqrt{3}}{2}}$

$\longrightarrow\bf{cos30^{\circ}=\dfrac{\sqrt{3}}{2}}$

$\longrightarrow\bf{cos60^{\circ}=\dfrac{1}{2}}$

Now, it is given that

$\longrightarrow\sf{tanA=\dfrac{1}{\sqrt{3}}}$

$\longrightarrow\sf{tanA=tan30^{\circ}}$

On comparing both the sides, we get

$\longrightarrow\sf{A=30^{\circ}}$

Also,

$\longrightarrow\sf{tanB=\sqrt{3}}$

$\longrightarrow\sf{tanB=tan60^{\circ}}$

On comparing both the sides, we get

$\longrightarrow\sf{B=60^{\circ}}$

Now,

$\longrightarrow\sf{sinA.cos B+cosA.sinB}$

On putting values of A and B in above expression, we get

$\longrightarrow\sf{sin(30^{\circ}).cos(60^{\circ})+cos(30^{\circ}).sin(60^{\circ})}$

$\longrightarrow\sf{\dfrac{1}{2}\times\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\times\dfrac{\sqrt{3}}{2}}$

$\longrightarrow\sf{\dfrac{1}{4}+\dfrac{3}{4}}$

$\longrightarrow\sf{\dfrac{1+3}{4}}$

$\longrightarrow\sf{\dfrac{4}{4}}$

$\longrightarrow\sf{1}$

Hence, the value of sinA.cosB+cosA.sinB is 1.

Answer 2

Answer:

Ans : 1

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