Question

sin A=,m sinB and tanA =n tanB then cos^2A = ?​

Answer 1

AnswEr :

⇝ sinA = m sinB

⇝ $\sf \dfrac{1}{\sin(B)}=\dfrac{m}{\sin(A)}$

⇝ $\sf \cosec(B) = \dfrac{m}{\sin(A)}$

⠀

⇝ tanA = n tanB

⇝ $\sf \dfrac{1}{ \tan(B) } = \dfrac{n}{ \tan(A) }$

⇝ $\sf \cot(B) = \dfrac{n}{\tan(A)}$

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• we know this trigonometric relation :

$\longrightarrow \sf\cosec^{2} (B) - \cot^{2} (B) = 1$

$\longrightarrow\bigg(\sf\dfrac{m}{\sin(B)}\bigg)^{2} - \bigg(\dfrac{n}{\tan(B)}\bigg)^{2} = 1$

$\longrightarrow\sf\dfrac{m^{2} }{\sin^{2} (B)}-\dfrac{n^{2} }{\tan^{2} (B)} = 1$

$\longrightarrow\sf\dfrac{m^{2} }{\sin^{2} (B)}-\dfrac{n^{2} }{ \frac{ \sin^{2} (B) }{ \cos^{2} (B) } } = 1$

$\longrightarrow\sf\dfrac{m^{2} }{\sin^{2} (B)}-\dfrac{n^{2} \cos^{2}(B)}{\sin^{2} (B)} = 1$

$\longrightarrow\sf\dfrac{m^{2} - n^{2} \cos^{2}(B)}{\sin^{2} (B)} = 1$

$\longrightarrow\sf m^{2} - n^{2} \cos^{2}(B) = \sin^{2} (B)$

$\longrightarrow\sf m^{2} - n^{2} \cos^{2}(B) = 1 - \cos^{2} (B)$

$\longrightarrow\sf m^{2} - 1 = n^{2} \cos^{2}(B) - \cos^{2} (B)$

$\longrightarrow\sf m^{2} - 1 = \cos^{2}(B)(n^{2} - 1)$

$\longrightarrow\boxed{\sf \cos^{2}(B) = \dfrac{m^{2} -1}{n^{2} - 1}}$

Answer 2

Given :----

• SinA = m sinB
• tanA = n tanB

To Find :----

• Cos²A

Formula used :----

• 1/sin@ = cosec@
• 1/tan@ = cot@
• cosec²@ - cot²@ = 1
• TanA = SinA/CosA
• Sin²A = 1 - Cos²A

Solution :-------

SinA = m sinB

→ 1/sinB = m/sinA

→ cosecB = m/sinA ----------------------Equation(1)

similarly,

tanA = n tanB

→ 1/tanB = n/tanA

→ cotB = n/tanA ----------------------Equation(2)

Now we know that,

cosec²B - cot²B = 1 -------------------Equation(3)

putting value of Equation (1) and Equation (2) in Equation (3) we get,

(m/SinA)² - (n/TanA)² = 1

putting value of tanA = sinA/cosA now,

➠ (m²/Sin²A) - (n²Cos²A/Sin²A) = 1

Taking LCM now,

➠ (m² - n²Cos²A) = Sin²A

putting value of sin²A in RHS now,

➠ (m² - n²Cos²A) = 1 - Cos²A

Taking LHS, RHS same side now,

➠ n²Cos²A - Cos²A = (m² - 1)

Taking cos²A common Now, from LHS

➠ Cos²A(n² - 1) = (m² - 1)

➠ Cos²A = (m²- 1) / (n² - 1)

(Hope it helps you)