Math, asked by MAYAKASHYAP5101, 1 year ago

4th part .

If tan A = ntanB , sinA = msinB ,
<i><b> cos²A = m²- 1 / ( n²-1 ) .

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Answers

Answered by siddhartharao77
7

Answer:

cos²A = (m² - 1)/(n² - 1)

Step-by-step explanation:

(4)

Given:

tan A = n tan B

⇒ n = (tan A/tan B)

⇒ tan B = tan A/n

⇒ cot B = n/tan A {∴ cotθ = 1/tanθ}


(ii)

sin A = m sin B

⇒ m = (sin A/sin B)

⇒ sin B = (sin A/m)

⇒ cosec B = (m/sin A) {∴ cosec θ = 1/sinθ}


LHS:

We know that cosec²B - cot²B = 1

⇒ (m/sinA)² - (n/tanA)² = 1

⇒ (m²/sin²A) - (n²/tan²A) = 1

⇒ (m²/sin²A) - (n² * cos²A)/sin²A) = 1

⇒ (m² - n²cos²A) = sin²A

⇒ m² - n²cos²A = (1 - cos²A)

⇒ m² - n²cos²A = 1 - cos²A

⇒ m² = 1 - cos²A + n² cos²A

⇒ m² = cos²A(n² - 1) + 1

⇒ m² - 1 = cos²A(n² - 1)

⇒ (m² - 1)/(n² - 1) = cos²A

RHS


Hope it helps!


nandhini18: is there any other format to answer this question
nandhini18: just a doubt
siddhartharao77: Yes..
MAYAKASHYAP5101: thankyou sooo much sir
siddhartharao77: welcome :-)
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