Question

IF
${m}^{2} + {m}^{ '2} + 2mm' \cos \theta = 1 {n }^{2} {n'}^{2} + 2nn' \cos \theta = 1 and \: mn + m'n' + (mn' + m'n) \cos \theta = 0 \: prove \: that \: {m}^{2} + {n}^{2} = { \cosec }^{2} \theta$
fast plz ​

Answer 1

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${m}^{2} + {m ' }^{2} + 2mm' \cos \theta = 1$

$\boxed {\mathtt{ \red S \blue O \orange L \purple U \green T \pink I \red ON}}$

${m'}^{2} + 2mm' \cos \theta = 1 - {m}^{2}$

$add \: {m}^{2} { \cos }^{2} \theta \: both \: sides$

${m ' }^{2} + 2mm' + {m}^{2} { \cos }^{2} \theta = 1 - {m}^{2} + {m}^{2} { \cos}^{2} \theta$

$(m' + {m}^{2} { \cos }^{2} \theta) = 1 - {m}^{2} { \sin}^{2} \theta...(i)$

${n}^{2} + {n' }^{2} + 2nn' \cos \theta = 1 \: \: \: \: \: \: (n' + n \cos \theta {)}^{2} = 1 - {n}^{2} { \sin }^{2} \theta.....(ii)$

$eq(i) \times (ii)$

$(m' + m \cos \theta {)}^{2} (n' + n \cos \theta {)}^{2} = (1 - {m}^{2} { \sin }^{2} \theta)(1 - {n}^{2} { \sin }^{2} \theta)(1 - {n}^{2} { \sin }^{2} \theta).....(iii)$

$(m' + m \cos \theta) (n' + n \cos \theta) = m'n' + (m'n + mn') \cos \theta + mn \: { \cos }^{2} \theta....(iv)$

$but \: mn + m'n' + (mn' + nm') \cos \theta = 1 \: in \: eq(iv)$

$(m' + m \cos \theta)(n'n \cos \theta) = 1 - mn + mn { \cos }^{2} \theta$

$= > 1 - mn(1 - { \cos }^{2} \theta)$

$= > 1 - mn \: { \sin }^{2} \theta...(v)$

$eq(v {)}^{2} \: we \: get$

$(m' + m \cos \theta)(n' + n \cos \theta) = 1 - mn + mn \: { \sin}^{2} \theta {)}^{2} .....(vi)$

$from \: eq(iii) \: and \: (vi)$

${m}^{2} {n}^{2} { \sin}^{2} \theta = (1 - {m}^{2} { \sin }^{2} \theta)(1 - {n}^{2} { \sin }^{2} \theta)$

$\cancel{{m}^{2} {n}^{2} { \sin }^{4} \theta }= 1 - {n}^{2} { \sin }^{2} \theta - {m}^{2} { \sin}^{2} \theta + \cancel{ {m}^{2} {n}^{2} { \sin }^{4} }$

${m}^{2} { \sin }^{2} \theta + {m}^{2} { \sin }^{2} \theta = 1$

${ \sin }^{2} \theta( {m}^{2} + {n}^{2} ) = 1$

${m}^{2} + {n}^{2} = \frac{1}{ { \sin }^{2} \theta }$

$\boxed{ {m}^{2} + {n}^{2} = { \cosec }^{2} \theta}$

HENCE PROVED.........

Answer 2

The given equation can be written as

$(m′+mcosθ2)+m2−m2cos2θ=1 </p><p></p><p>$

Similarly

$n′+ncosθ)2=1−n2sin2θ</p><p></p><p>$

Now

$(m′+mcosθ)=(n′+ncosθ)</p><p></p><p>$

$=−mn+mncos2θ </p><p></p><p>$

By Given Relation

$=−mn(1−cos2θ)=−mnsin2θ</p><p></p><p>$

Now squaring both sides, we get

$or (m′+mcosθ)2(n′+ncosθ)2=m2n2sin4θ</p><p></p><p>$

Hence substituting from (1) and (2) in (3), we get

$(1−m2sin2θ)(1−n2sin2θ)=m2n2sin4θ</p><p></p><p>$

$or =(m2+n2)sin2θ=1 i.e. =m2+n2=csc2θ</p><p></p><p>$

Hence,We proved this⬆⬆