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[tex]\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\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}\end{gathered}\end{gathered}\end{gathered}\end{gathered}[/tex]
Eᴠᴀʟᴜᴀᴛᴇ :-
I) sɪɴ60°ᴄᴏs30°+sɪɴ30°ᴄᴏs60°
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[tex]\huge\color{black}\boxed{\colorbox{aqua}{Solution}}[/tex]
In the expression sin(60°)cos(30°) + sin(30°)cos(60°), we have two trigonometric functions, sine and cosine, and angles of 60° and 30°.
First, let's evaluate sin(60°)cos(30°). The sine of 60° is √3/2, and the cosine of 30° is √3/2. So, sin(60°)cos(30°) becomes (√3/2)(√3/2) = 3/4.
Next, let's evaluate sin(30°)cos(60°). The sine of 30° is 1/2, and the cosine of 60° is 1/2. So, sin(30°)cos(60°) becomes (1/2)(1/2) = 1/4.
Now, we can substitute these values back into the original expression: 3/4 + 1/4 = 4/4 = 1.
Therefore, the value of sin(60°)cos(30°) + sin(30°)cos(60°) is 1.
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Verified answer
We know that,
Substituting all the values, we get :
→ (√3/2) × (√3/2) + (1/2) × (1/2)
→ (√3 × √3)/(2 × 2) + 1/(2 × 2)
→ 3/4 + 1/4
→ (3 + 1)/4
→ 4/4
→ 1 ... [Ans]
Therefore, the value of the above given expression = 1.
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I hope you got it clear, doe :)
Didn't use latex cuz that thing's too boring to do. :/ [still I hope you understand]
–adeline :))