Question

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

Answer 1

Answer:

1

Step-by-step explanation:

tan A = 1/√3 .... A = 30 degree

tan B = √3 .... B = 60 degree

sin A.cosB+cosA.sinB

1/2 * 1/2 + √3/2 * √3/2

=1

Answer 2

Given:-

• $\sf{tanA=\dfrac{1}{\sqrt{3}}}$

• $\sf{tanB=\sqrt{3}}$

To Find:-

• Value of sin A.cos B+cos A.sin B

Notes :-

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

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

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

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

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

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

Solution:-

Given that,

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

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

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

Again,

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

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

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

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Now,put the value of A and B.

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

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

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

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

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

$\longrightarrow\sf{1}$

Hence,

the value of sinA.cosB+cosA.sinB is = 1

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