Question

if A,B are supplementary angles then sin^2 A + sin^2​

Answer 1

Step-by-step explanation:

Given,

A, B are supplementary angles.

To Find :-

sin²A + sin²B

Solution :-

A, B are supplementary angles.

→ A + B = 180°

B = 180° - A

sin²A + sin²B

= sin²A + sin²(180° - A)

= sin²A + sin²A

[sin²(180°-A) is second quadrant so "sin" is positive ]

= 2sin²A

sin²A + sin²B = 2sin²A

Know More :-

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

The Answer is 1.

I know this because A,B are supplementary angles.

And A+B = 180 degrees And B=180 degrees -A

(sum of these supplementary angles are 180 degrees)

Now,  Cos A+ Sin B (here, in question something is missing. so i assumed it as

Cos A-E sin  B)  =Cos A+ Sin (180-A)

                       =Cos A+ Sin A

(Sin(180-0)=Sin 0)

=1

 =                     thank you for your time! plz mark me brainliest!  spent 33 mins on this!