Question

guys please help me i, question is in attachment (class X Maths).
Please guys help me ​

Answer 1

Given :-

• tanA = n*tanB
• sinA = m*sinB

To Prove :-

• cos²A = (m² - 1)/(n² - 1)

Solution :-

→ tanA = n*tanB

Cross - Multiplying ,

→ 1/tanB = n/tanA

Now, we know That 1/tan@ = cot@ .

So,

→ cotB = n/tanA ----------------- Equation (1).

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Similarly,

→ sinA = m*sinB

→ 1/sinB = m/sinA

Now, we know That 1/sin@ = cosec@ .

Now, we know That 1/sin@ = cosec@ .So,

→ cosecB = m/sinA -------------- Equation (2).

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Now, we know That, cose²@ - cot²@ = 1.

Putting Both Values from Equation (1) & (2) now, we get,

→ cosec²B - cot²B = 1

Or,

→ (m/sinA)² - (n/tanA)² = 1

Now putting tanA = (sinA /cosA) we get,

→ (m²/sin²A) - (n²cos²A/sin²A) = 1

Taking LCM,

→ (m² - n²cos²A) /sin²A = 1

→ (m² - n²cos²A) = sin²A

Now, putting sin²A = (1 - cos²A) in RHS, we get,

→ (m² - n²cos²A) = ( 1 - cos²A)

→ n²cos²A - cos²A = (m² - 1)

Taking cos²A common From LHS,

→ cos²A(n² - 1) = (m² - 1)

→ cos²A = (m² - 1) / (n² - 1) = Hence, Proved .

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Answer 2

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