Question

Please help me with this question​

Answer 1

$\;\underbrace{\underline{\sf{Understanding\;the\;Concept}}}$

Here the concept of Trignometric Identities have been used. We see that we are given a equation to prove LHS = RHS.

★ Solution :-

Given,

$\\\;\tt{\odot\;\;a\:\cos^{3}\theta\;+\;3a\cos\theta\:\sin^{2}\theta\;=\;\bf{m}}$

$\\\;\tt{\odot\;\;a\:\sin^{3}\theta\;+\;3a\cos^{2}\theta\:\sin\theta\;=\;\bf{n}}$

~ For the value of (m + n) ::

This is given as,

$\\\;\sf{\rightarrow\;\;(m\:+\:n)\;=\;\bf{(a\:\cos^{3}\theta\;+\;3a\cos\theta\:\sin^{2}\theta)\;+\;(a\:\sin^{3}\theta\;+\;3a\cos^{2}\theta\:\sin\theta)}}$

On adding we get,

$\\\;\sf{\rightarrow\;\;(m\:+\:n)\;=\;\bf{a\:\cos^{3}\theta\;+\;3a\cos\theta\:\sin^{2}\theta\;+\;a\:\sin^{3}\theta\;+\;3a\cos^{2}\theta\:\sin\theta}}$

Taking out the common terms, we get

$\\\;\sf{\rightarrow\;\;(m\:+\:n)\;=\;\bf{a(\cos^{3}\theta\:+\:\sin^{3}\theta)\;+\;3a\cos\theta\:\sin\theta(\cos\theta\:+\;\sin\theta)}}$

This forms similar identity like,

✒ (a + b)³ = a³ + b³ + 3ab(a + b)

• Here a = cos θ and b = sin θ

Now applying this, we get

$\\\;\bf{\rightarrow\;\;(m\:+\:n)\;=\;\bf{\orange{a(\cos\theta\:+\:\sin\theta)^{3}}}}$

~ For the value of (m - n) ::

This is given as,

$\\\;\sf{\rightarrow\;\;(m\:-\:n)\;=\;\bf{(a\:\cos^{3}\theta\;+\;3a\cos\theta\:\sin^{2}\theta)\;-\;(a\:\sin^{3}\theta\;+\;3a\cos^{2}\theta\:\sin\theta)}}$

On opening the bracket, we get

$\\\;\sf{\rightarrow\;\;(m\:-\:n)\;=\;\bf{a\:\cos^{3}\theta\;+\;3a\cos\theta\:\sin^{2}\theta\;-\;a\:\sin^{3}\theta\;-\;3a\cos^{2}\theta\:\sin\theta}}$

Taking out the common terms, we get

$\\\;\sf{\rightarrow\;\;(m\:-\:n)\;=\;\bf{a(\cos^{3}\theta\:-\:\sin^{3}\theta)\;-\;3a\cos\theta\:\sin\theta(\cos\theta\:-\;\sin\theta)}}$

This forms a identity similar like,

✒ (a - b)³ = a³ - b³ - 3ab(a - b)

• Here a = cos θ and b = sin θ

Now applying this, we get

$\\\;\bf{\rightarrow\;\;(m\:-\:n)\;=\;\bf{\orange{a(\cos\theta\:-\:\sin\theta)^{3}}}}$

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~ For proving the given equation ::

We are given to prove that,

$\\\;\bf{\mapsto\;\;(m\:+\:n)^{2/3}\;+\;(m\:-\:n)^{2/3}\;=\;2\:a^{2/3}}$

Now let's take LHS first,

$\\\;\bf{\Longrightarrow\;\;L.H.S.\;=\;\bf{(m\:+\:n)^{2/3}\;+\;(m\:-\:n)^{2/3}\;=\;2\:a^{2/3}}}$

Now by applying values that we got above, we get

$\\\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{\bigg(a(\cos\theta\:+\:\sin\theta)^{3}\bigg)^{2/3}\;+\;\bigg(a(\cos\theta\:-\:\sin\theta)^{3}\bigg)^{2/3}}}$

$\\\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{a^{2/3}\bigg(\cos\theta\:+\:\sin\theta)^{3}\bigg)^{2/3}\;+\;a^{2/3}\bigg(\cos\theta\:-\:\sin\theta)^{3}\bigg)^{2/3}}}$

We see that the term a with its exponent is common in both, so let's take it common

$\\\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{a^{2/3}\bigg((\cos\theta\:+\:\sin\theta)^{3})^{2/3}\;+\;(\cos\theta\:-\:\sin\theta)^{3})^{2/3}\bigg)}}$

Now cancelling the powers by laws of exponents,

$\\\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{a^{2/3}\bigg((\cos\theta\:+\:\sin\theta)^{2}\;+\;(\cos\theta\:-\:\sin\theta)^{2}\bigg)}}$

We know that,

✒ (a + b)² = a² + b² + 2ab

• Here a = cos θ and b = sin θ

By applying values, we get

$\\\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{a^{2/3}\bigg((\cos^{2}\theta\:+\:\sin^{2}\theta\:+\:2\cos\theta\sin\theta)\;+\;(\cos\theta\:-\:\sin\theta)^{2}\bigg)}}$

We know that,

✒ (a - b)² = a² + b² - 2ab

• Here a = cos θ and b = sin θ

$\\\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{a^{2/3}\bigg((\cos^{2}\theta\:+\:\sin^{2}\theta\:+\:2\cos\theta\sin\theta)\;+\;(\cos^{2}\theta\:+\:\sin^{2}\theta\:-\:2\cos\theta\sin\theta)\bigg)}}$

Now opening the bracket, we get

$\\\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{a^{2/3}\bigg(\cos^{2}\theta\:+\:\sin^{2}\theta\:+\:2\cos\theta\sin\theta\;+\;\cos^{2}\theta\:+\:\sin^{2}\theta\:-\:2\cos\theta\sin\theta\bigg)}}$

Cancelling the like terms, we get

$\\\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{a^{2/3}\bigg(\cos^{2}\theta\:+\:\sin^{2}\theta\;+\;\cos^{2}\theta\:+\:\sin^{2}\theta\bigg)}}$

Now adding the like terms, we get

$\\\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{a^{2/3}\bigg(2\cos^{2}\theta\;+\;2\sin^{2}\theta\bigg)}}$

Now taking 2 as common, we get

$\\\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{2a^{2/3}\bigg(\cos^{2}\theta\;+\;\sin^{2}\theta\bigg)}}$

$\\\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{2a^{2/3}\bigg(\sin^{2}\theta\;+\;\cos^{2}\theta\bigg)}}$

We already know that,

$\\\;\tt{\rightarrow\;\;\sin^{2}\theta\;+\;\cos^{2}\theta\;=\;1}$

By applying this here, we get

$\\\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{2a^{2/3}\bigg(1\bigg)}}$

$\\\;\bf{\Longrightarrow\;\;L.H.S.\;=\;\bf{\blue{2a^{2/3}}}}$

We also know that,

$\\\;\bf{\Longrightarrow\;\;R.H.S.\;=\;\bf{\blue{2a^{2/3}}}}$

From this we get,

$\\\;\bf{\red{\Longrightarrow\;\;L.H.S.\;=\;R.H.S.\;=\;\bf{2a^{2/3}}}}$

$\\\;\qquad\qquad\boxed{\underline{\tt{\purple{Hence,\;\:Proved}}}}$

Answer 2

$\begin{gathered}\begin{gathered}\bf Given : - \begin{cases} &\sf{\begin{gathered}\\\;\sf{\;a\:\cos^{3}\theta\;+\;3a\cos\theta\:\sin^{2}\theta\;=\;\sf{m}}\end{gathered}} \\ &\sf{\begin{gathered}\\\;\sf{\;a\:\sin^{3}\theta\;+\;3a\cos^{2}\theta\:\sin\theta\;=\;\sf{n}}\end{gathered}} \end{cases}\end{gathered}\end{gathered}$

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$\begin{gathered}\begin{gathered}\bf To \: prove :- \begin{cases} &\bf{ {(m + n)}^{ \frac{2}{3} } + {(m - n)}^{ \frac{2}{3} } = 2 {a}^{ \frac{2}{3} } } \end{cases}\end{gathered}\end{gathered}$

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$\begin{gathered}\Large{\bold{\pink{\underline{Formula \: Used \::}}}} \end{gathered}$

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$\bf \:  ⟼ (a + b)³ = a³ + b³ + 3ab(a + b) \\ \bf \:  ⟼ (a + b)³ = a³ + b³ + 3ab(a + b)$

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$\large\underline\purple{\bold{Solution :- }}$

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$\bf \: \underline{ ❥︎ Step :- 1}$

❥︎ Consider

$\sf \:  ⟼{m + n}$

$\sf \:  = a {cos}^{3} \theta + 3a {sin}^{2} \theta \: cos\theta + a {sin}^{3} \theta + 3a {cos}^{2} \theta \: sin\theta$

$\sf \:  = a ({cos}^{3} \theta + 3 {sin}^{2} \theta \: cos\theta + {sin}^{3} \theta + 3 {cos}^{2} \theta \: sin\theta)$

$\sf \:  = a( {cos}^{3} \theta + {sin}^{3} \theta + 3sin\theta \: cos\theta \: (sin\theta \: + cos\theta \: ))$

$\sf \:  = a {(sin\theta \: + cos\theta \:) }^{3}$

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$\bf \: \underline{ ❥︎ Step :- 2}$

$\sf \:  ⟼m - n$

$\sf \:  = (a {cos}^{3} \theta + 3a {sin}^{2} \theta \: cos\theta) - ( a {sin}^{3} \theta + 3a {cos}^{2} \theta \: sin\theta)$

$\sf \:  = a ({cos}^{3} \theta + 3a {sin}^{2} \theta \: cos\theta - a {sin}^{3} \theta - 3a {cos}^{2} \theta \: sin\theta)$

$\sf \:  = a( {cos}^{3} \theta - {sin}^{3} \theta - 3sin\theta \: cos\theta \: (cos\theta \: - sin\theta \: ))$

$\sf \:  = a{(cos\theta \: - sin\theta \:)}^{3}$

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$\bf \: \underline{ ❥︎ Step :- 3}$

❥︎ Now, Consider

$\sf \:  ⟼{(m + n)}^{ \frac{2}{3} } = { {(a(cos\theta \: + sin\theta \:))}^{3 \times \frac{2}{3} } }$

$\sf \:  ⟼{(m + n)}^{ \frac{2}{3} } = {a}^{ \frac{2}{3} } {(cos\theta \: + sin\theta \:)}^{2}$

$\sf \:  ⟼{(m + n)}^{ \frac{2}{3} } = {a}^{ \frac{2}{3} } ( {cos}^{2} \theta \: + {sin}^{2} \theta \: + 2sin\theta \:cos\theta \:)$

$\sf \:  ⟼{(m + n)}^{ \frac{2}{3} } = {a}^{ \frac{2}{3} } (1 + 2sin\theta \:cos\theta \:)\bf \:  ⟼ (1)$

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$\bf \: \underline{ ❥︎ Step :- 4}$

❥︎ Now, Consider

$\sf \:  ⟼{(m - n)}^{ \frac{2}{3} } = { {(a(cos\theta \: - sin\theta \:))}^{3 \times \frac{2}{3} } }$

$\sf \:  ⟼{(m - n)}^{ \frac{2}{3} } = {a}^{ \frac{2}{3} } {(cos\theta \: - sin\theta \:)}^{2}$

$\sf \:  ⟼{(m - n)}^{ \frac{2}{3} } = {a}^{ \frac{2}{3} } ( {cos}^{2} \theta \: + {sin}^{2} \theta \: - 2sin\theta \:cos\theta \:)$

$\sf \:  ⟼{(m - n)}^{ \frac{2}{3} } = {a}^{ \frac{2}{3} } (1 - 2sin\theta \:cos\theta \:)\bf \:  ⟼ (2)$

$\bf \: \underline{ ❥︎ Step :- 5}$

❥︎ Consider LHS,

$\bf \:{(m + n)}^{ \frac{2}{3} } + {(m - n)}^{ \frac{2}{3} }$

$\sf \:  = {a}^{ \frac{2}{3} } (1 + 2sin\theta \:cos\theta \:) + {a}^{ \frac{2}{3} } (1 - 2sin\theta \:cos\theta \:)$

$\sf \:  = {a}^{ \frac{2}{3} } \bigg( (1 + 2sin\theta \:cos\theta \:) + (1 - 2sin\theta \:cos\theta \:) \bigg)$

$\sf \:  = {a}^{ \frac{2}{3} } \bigg( 1 + \cancel{2sin\theta \:cos\theta \:} + 1 - \cancel{2sin\theta \:cos\theta \:} \bigg)$

$\sf \:  = 2 {a}^{ \frac{2}{3} }$

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$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$

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$\large \red{\bf \: ⟼ Explore \: more } ✍$

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Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

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Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

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Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

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Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1

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