Question

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$\sin(25) \cos(65) + \cos(25) \sin(65)$
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Answer 1

as sin theta=cos(90-theta)

so sin 25=cos(90-25)=cos 65

likewise

as cos theta=sin(90-theta)

so cos 25=sin(90-25)=sin 65

therefore

sin(25)cos(65) + cos(25)sin(65)

cos(65)cos(65) + sin(65)sin(65)

cos²(65) + sin²(65) = 1

from identity sin²theta+cos²theta=1

Answer 2

$\sf \Large \underbrace{ \underline{Explaination}}$

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$\small\text{=>Sin(25°)Cos(65°)+Cos(25°)Sin(65°)}$

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$\small\text{=>Sin(25°)Cos(90°-25°)+Cos(25°)Sin(90°-25°)}$

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$\small\text{=>Sin(25°)Sin(25°)+Cos(25°)Cos(25°)}$

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$\small\text{=>Sin²(25°)+Cos²(25°)}$

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$\sf =>1$ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎[•°•Sin²@+Cos²@=1]

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$\sf \large \underbrace{ \underline{Trigonometry\: formula}}$

$\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ \: \: 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1 \end{minipage}}$

Trigonometry ratio:

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

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