Question

guys please help me question is in attachment.Maths class 10 ​

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@ .So,

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

______________________

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

---------------METHOD(1)-------------------

GIVEN :-

• n = tanA/tanB

• m = sinA/ sinB

TO Prove :-

cos²A

Proof :-

=> R.H.S = m²-1/n²-1

=> sin²A/sin²B -1 / tan²B/tan²B -1

=> sin²A - sin²B/sin²B / tan²A - tan²B/ tan²B

=>.°. sin²A - sin²B/sin²B / tan²A - tan²B/ tan²B = sin²B / cos²B

=> sin²A - sin²B / cos²B(tan²A - tan²B)

=> 1-cos²A -1 + cos²B / cos²B(sec²A-1-sec²+1)

=> cos²B - cos²A / cos²B(1/cos²A - 1/cos²B)

=> cos²B - cos²A / cos²B (cos²B - cos²A) /cos²A.cos²B.

=>L.H.S = Cos²A

=> .°. R.H.S = cos²A = L. H. S

---------------METHOD(2)-------------------

solution :-

=> tan²A = n tanB = tanB = 1/n tanA = cot²A = n/tanA

=> sinA = m sinB = 1/m sinA = cosecB = m/sinA.

=> putting the value of CotB and cosecB in cosec²B - cot²B = 1 , we get,

=> m²/sin²A - n²/tan²A = 1

=> m²/sin²A - n²cos²A/sin²A = 1

=> m²-n²cos²A / sin²A= 1

=> m²-n² cos²A = sin²A

=> m²-n² cos²A = 1 - sin²A [°.°sin²A =1 -cos²A]

=> m²-1 = n² cos²A - cos²A

=> m² - 1 = (n²-1) cos²A

=> .°. m²-1/n²-1 = cos²A.