Question

if sin A=1/√2 and cot B=1 .Prove that sin(A+B)=1 where A &B both are acute angles

Answer 1

Angle are acute
SinA=1/√2
=sin45°
A=45°
CotB=1
=cot45°
B=45°
Sin(A+B)=sin(45°+45°)
=sin90°
=1
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Answer 2

sin(A+B)=1

Step-by-step explanation:

Sin A = $\frac{1}{\sqrt{2} }$                             Cot B = 1

 A = $Sin^{-1}$ $\frac{1}{\sqrt{2} }$                            B = $cot^{-1}$ 1

  A = 45°                                     B = 45°  

Sin(A + B)

= Sin(45° + 45°)

= Sin(90°)

= 1

Learn more: Trigonometry

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