13. The sender transmits the message while the receiver accepts the message.
The most important factor in communication is the good delivery of the message to ensure better
understanding which will result in a satisfying answer. Moreover, the tool in communication is as
important as the message being sent because it is the channel by which the message is delivered.
a. Communication is a process by which a message is transmitted.
b. Communication has three important elements: the sender, the receiver, and the medium.
c. Communication happens when a sender gives information to the receiver.
Answers
Answer:
Step-by-step explanation:
\LARGE{\bf{\underline{\underline\color{blue}{GIVEN:-}}}}
GIVEN:−
\sf \bullet \ \ \dfrac{(1+sinA-cosA)^2}{(1+sinA+cosA)^2} = \dfrac{1-cos}{1+cos}∙
(1+sinA+cosA)
2
(1+sinA−cosA)
2
=
1+cos
1−cos
\LARGE{\bf{\underline{\underline\color{blue}{SOLUTION:-}}}}
SOLUTION:−
LHS:
\sf \to \dfrac{(1+sinA-cosA)^2}{(1+sinA+cosA)^2}→→
(1+sinA+cosA)
2
(1+sinA−cosA)
2
→
Expand the fractions using (a+b+c)²=a²+b²+c²+2ab+2bc+2ca.
\sf \to \dfrac{(cos^2-2sincos+sin^2-2cos+2sin+1)}{(cos^2+2sincos+sin^2+2cos+2sin+1)}→→
(cos
2
+2sincos+sin
2
+2cos+2sin+1)
(cos
2
−2sincos+sin
2
−2cos+2sin+1)
→
Rearrange the terms.
\sf \to \dfrac{(cos^2+sin^2-2sincos-2cos+2sin+1)}{(cos^2+sin^2+2sincos+2cos+2sin+1)}→→
(cos
2
+sin
2
+2sincos+2cos+2sin+1)
(cos
2
+sin
2
−2sincos−2cos+2sin+1)
→
We know that cos²A+sin²A=1.
\sf \to \dfrac{1-2sincos-2cos}{2sin+1}→→
2sin+1
1−2sincos−2cos
→
Now here, take -2cos common from the numerator and +2cos common from the denominator.
\sf \to \dfrac{1-2cos(sin+2)}{2sin+1}→→
2sin+1
1−2cos(sin+2)
→
Now, rearrange the terms, add 1 and 1 and take 2 common.
\to\sf\dfrac{1+1+2sin-2cos}{sin+1}→→
sin+1
1+1+2sin−2cos
→
\to\sf\dfrac{2+2sin-2cos}{sin+1}→→
sin+1
2+2sin−2cos
→
Take 2 common.
\to \sf \dfrac{ 2(1+sin) -2cos(sin+1) }{ 2(1+sin) + 2cos(sin +1 ) }→→
2(1+sin)+2cos(sin+1)
2(1+sin)−2cos(sin+1)
→
\to \sf{\red{\dfrac{1-cosA}{1+cosA} }}→ < /p > < p >→
1+cosA
1−cosA
→</p><p>
LHS=RHS.
HENCE PROVED!
FUNDAMENTAL TRIGONOMETRIC RATIOS:
\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\substack{\displaystyle \sf sin^2 \theta+cos^2 \theta = 1 \\\\ \displaystyle \sf 1+cot^2 \theta=cosec^2 \theta \\\\ \displaystyle \sf 1+tan^2 \theta=sec^2 \theta}}\end{gathered}\end{gathered} \end{gathered}
sin
2
θ+cos
2
θ=1
1+cot
2
θ=cosec
2
θ
1+tan
2
θ=sec
2
θ
T-RATIOS:
\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 Not \: De fined \\ \\ \rm cosec A & \rm Not \: De fined & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm Not \: De fined \\ \\ \rm cot A & \rm Not \: De fined & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}}}\end{gathered}\end{gathered} \end{gathered}
∠A
sinA
cosA
tanA
cosecA
secA
cotA
0
∘
0
1
0
NotDefined
1
NotDefined
30
∘
2
1
2
3
3
1
2
3
2
3
45
∘
2
1
2
1
1
2
2
1
60
∘
2
3
2
1
3
3
2
2
3
1
90
∘
1
0
NotDefined
1
NotDefined
0