Question

2. Prove that:
(i) sin(n+1)A}.sin(n+2)A)
+cos (n(n+1)4) cos{(n+2)A) =cos A​

Answer 1

Answer:

$\large\underline{\underline{\rm{\red{❒Given:-}}}}$

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• 2. Prove that:
• (i) sin(n+1)A}.sin(n+2)A)+cos (n(n+1)4) cos{(n+2)A) =cos A

${\underline{\color{cyan}{\rule{300pt}{9pt}}}}$

❒To Find:-

• Prove

$\large\underline{\underline{\rm{\orange{ ❒Concept:- }}}}$

• Trigonometric Functions of Sum and Difference of Angles

$\large\underline{\underline{\rm{\red{❒Solution:-}}}}$

${ \sf{\underline{ \longmapsto{ \sf \red{L.H.S}.= sin \: x \: sin \: y + cos \: X \: cos \: y}}}}$

${ \sf{\underline{ \longmapsto{ \sf \red{L.H.S}= cos \: x \: cos \: \: y + sin \: x \: sin \: y}}}}$

${ \sf{\underline{ \longmapsto{ \sf \red{L.H.S}=cos(x - y)}}}}$

${ \sf{\underline{ \longmapsto{ \sf \red{L.H.S}= cos [(n + 1)A - (n + 2)A]}}}}$

${ \sf{\underline{ \longmapsto{ \sf \red{L.H.S}= cos (nA + A - NA - 2A)}}}}$

${ \sf{\underline{ \longmapsto{ \sf \red{L.H.S}=cos(-A)}}}}$

${ \sf{\underline{ \boxed{ \pink{ \longmapsto{ \sf \bold{L.H.S}= cos A}}}}}}$

= R.H.S.

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More Information:-

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

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