Question

Find the acute angle between the pair of lines represented by ( x cos alha - y sin alpha)^2 = ( x^2 + y^2) sin^2 alpha

Answer 1

Answer:

(xcos@ - ysin@)² = (x²+y²)sin²@

=> x²cos²@ - 2xycos@sin@ + y²sin²@ = x²sin²@ + y²sin²@

=> x²cos²@ - x²sin²@ - 2xycos@sin@ = 0

=> x²(cos²@-sin²@) - xy(2sin@cos@)=0

=> x²cos2@ - xysin2@ = 0

=> a = cos2@ , b =0 , 2h = -sin2@=>h= -sin2@/2

=> tanβ = 2√(h²-ab) /(a+b)

=> tanβ = 2√((-sin2@/2)²-0) / cos2@

=> tanβ = 2√(sin²2@/4) / cos2@

=>tanβ = 2sin2@/2/cos2@

=> tanβ = sin2@/cos2@

=>tanβ = tan2@

=> β=2@

Answer 2

Step-by-step explanation:

$(x \cos\alpha - y \sin \alpha ) ^{2} = (x ^{2} + y ^{2} ) \sin( ^{2} ) \alpha \\ {x}^{2} { \cos}^{2} \alpha + {y}^{2} { \sin }^{2} \alpha - 2xy \cos\alpha \sin\alpha = {x}^{2} { \sin}^{2} \alpha + {y}^{2} { \sin}^{2} \alpha \\ {x}^{2} ( { \cos }^{2} \alpha - { \sin }^{2} \alpha ) - 2xy \cos\alpha \sin\alpha = 0 \\ {x}^{2} \cos(2 \alpha ) - xy \sin(2 \alpha ) = 0 \\ a = \cos(2 \alpha ) \: \\ b = 0 \\ 2h = - \sin(2 \alpha ) \\ let \: \gamma n \: be \: the \: angle \: between \: the \: lines. \: then \\ tan \gamma = 2 \sqrt{ { \sin}^{2} } \alpha \div 4 - 0 \: by \: | \cos(2 \alpha ) \: + 0| \: = \tan(2 \alpha ) \\ \gamma = 2 \alpha$

hope its help uhh

Mark me brainlist...