Question

প্ৰমাণ কৰা যে (i)Sin45°=1/√2
(ii)Sin(90°-A)=Cos A
(জ্যামিতিব্দ ভাৱে)​

Answer 1

Given:

$\bold{(i) \sin 45^{\circ} = \frac{1}{\sqrt{2}}}\\\\\bold{(ii) \sin(90^{\circ}-A) = \cos A}$

To find:

Prove.

Solution:

In the trigonometry $(\sin (90^{\circ}- \theta)= \cos \theta \ \ that's \ why \ \ \sin (90^{\circ}- A)= \cos A )$ and $\sin 45^{\circ} = \frac{1}{\sqrt{2}}$ that's why both the given equation is true.

Answer 2

Answer:

what a question???!?!?