Question

if tan theta + sin theta is equal to M and 10 theta minus sin theta is equal to N show that n square minus n square is equal to 4 under root a man​

Answer 1

correct Question :- If tanA+sinA=m, and tanA-sinA=n, show that m²-n² = 4√mn ..

Solution :-

Taking LHS, First :-

→ m² - n²

→ (tanA+sinA)²-(tanA-sinA)²

using (a+b)² - (a-b)²= 4ab we get,

→ 4*tanA*sinA

_________________

Taking RHS now :-

→ 4√(m * n )

→ 4√(tanA+sinA)(tanA-sinA)

using (a + b)(a - b) = a² - b² now,

→ 4√(tan²A-sin²A)

using tanA = (sinA/cosA) now,

→ 4√[(sin²A/cos²A) - sin²A]

→ 4√[ (sin²A - sin²A cos²A)/cos²A ]

→ 4√[sin²A(1 - cos²A) / cos²A ]

using (1 - cos²A) = sinA now,

→ 4√[(sin²A*sin²A)/cos²A]

→ 4√[(sin⁴A)/(cos²A)]

→ 4 * (sin²A/cosA)

→ 4 * (sinA/cosA) * sinA

→ 4 * tanA * sinA = LHS

✪✪ Hence Proved ✪✪

_________________

Answer 2

Correct question:-

If tan∅ + sin∅ is equal to M and tan∅ - sin∅ is equal to N. Show that : M² - N² = 4√MN

Given :-

• tan∅ + sin∅ = M
• tan∅ - sin∅ = N

To show :-

• M² - N² = 4√MN

Proof:-

First we find : M²

$\rm M^2 = (\tan \theta + \sin \theta)^2 \\\\\qquad\scriptsize\sf{\big [Using \: identity : (a + b)^2 = a^2 + 2ab + b^2 \big ]}\\\\\longrightarrow\rm \tan^2 \theta + 2 \tan \theta \sin \theta + \sin^2 \theta \dots \dots (i)$

Now let's find N²

$\rm N^2 = (\tan \theta - \sin \theta)^2 \\\\\qquad\scriptsize\sf{\big[ Using \: identity : (a - b)^2 = a^2 - 2ab + b^2 \big ]}\\\\\longrightarrow\rm \tan^2 \theta - 2 \tan \theta \sin \theta + \sin^2 \theta \dots \dots (ii)$

Now we will find (i) - (ii) :

$\rm M^2 = \tan^2 \theta + 2 \tan \theta \sin \theta + \sin^2 \theta \\\\\rm N^2 = \tan^2 \theta - 2 \tan \theta \sin \theta + \sin^2 \theta \\\\\quad\sf \: (-) \qquad (+) \qquad \qquad (-)$

$\rule{200}1$

$\longrightarrow\sf\green{ M^2 - N^2 = 4 \tan \theta \sin \theta \quad[L.H.S.]}$

$\rule{200}{1}$

Now we will find MN

→ MN = (tan∅ + sin∅)(tan∅ - sin∅)

→ MN = tan²∅ - tan∅sin∅ + tan∅sin∅ - sin²∅

→ MN = tan²∅ - sin²∅

[∵ tan²∅ = sin²∅/cos²∅ ]

→ MN = (sin²∅/cos²∅) - sin²∅

→ MN = (sin²∅ - sin²∅cos²∅)/cos²∅

→ MN = [sin²∅(1 - cos²∅)]/cos²∅

[∵ (1 - cos²∅) = sin²∅]

→ MN = [sin²∅ × sin²∅]/cos²∅

→ MN = (sin²∅/cos²∅) × sin²∅

[∵ sin²∅/cos²∅ = tan²∅ ]

→ MN = tan²∅ sin²∅

Now we will find 4√MN

→ 4√MN = 4√(tan²∅ sin²∅)

→ R.H.S. = 4 tan∅ sin∅

Therefore,

L.H.S.= R.H.S.= 4 tan∅ sin∅ [Showed]