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If cos A=ncos...

Question

If $cos A = n cos B$ and $sin A = m sin B$ , then $(m^{2} - n^{2}) {sin}^{2} B =$

A

1

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B

1 + n^{2}

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C

n^{2}

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D

1 - n^{2}

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Solution

The correct option is D $1 - n^{2}$
Given
$cos A = n cos B \Rightarrow {cos}^{2} A = n^{2} {cos}^{2} B \dots (1)$
and
$sin A = m sin B \Rightarrow {sin}^{2} A = m^{2} {sin}^{2} B \dots (2)$
adding both the equation we have
$1 = m^{2} {sin}^{2} B + n^{2} {cos}^{2} B m^{2} {sin}^{2} B + n^{2} - n^{2} {sin}^{2} B = 1 (m^{2} - n^{2}) {sin}^{2} B = 1 - n^{2}$

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Similar questions

cos A = n cos B

sin A = m sin B

(m^{2} - n^{2}) {sin}^{2} B =

m = \frac{sin A}{sin B}, n = \frac{cos A}{cos B}

, then prove that

(m^{2} - n^{2}) {sin}^{2} B = 1 - n^{2}

Q. If tan A = n tan B and sin A = m sin B, prove that cos²A =

\frac{(m^{2} - 1)}{n^{2} - 1} .

m = cos A - sin A

n = cos A + sin A

\frac{m^{2} + n^{2}}{m^{2} - n^{2}} = - \frac{1}{2} sec A . cos e c A = - \frac{(cot A + tan A)}{2}

\frac{c o s B}{s i n A} = n a n d \frac{c o s B}{c o s A} = m,

(m^{2} + n^{2}) c o s^{2} A = n^{2} .

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