IFfast plz

1.	IFfast plz
Answer» $\boxed {\mathbb{ \red{G} \blue{I} \pink{V} \purple{E} \orange{ N}}}$ ${m}^{2} + {m ' }^{2} + 2mm' \cos \theta = 1$ $\boxed {\mathtt{ \red S \blue O \orange L \purple U \green T \pink I \red ON}}$ ${m'}^{2} + 2mm' \cos \theta = 1 - {m}^{2}$ $add \: {m}^{2} { \cos }^{2} \theta \: both \: sides$ ${m ' }^{2} + 2mm' + {m}^{2} { \cos }^{2} \theta = 1 - {m}^{2} + {m}^{2} { \cos}^{2} \theta$ $(m' + {m}^{2} { \cos }^{2} \theta) = 1 - {m}^{2} { \sin}^{2} \theta...(i)$ ${n}^{2} + {n' }^{2} + 2nn' \cos \theta = 1 \: \: \: \: \: \: (n' + n \cos \theta {)}^{2} = 1 - {n}^{2} { \sin }^{2} \theta.....(ii)$ $eq(i) \times (ii)$ $(m' + m \cos \theta {)}^{2} (n' + n \cos \theta {)}^{2} = (1 - {m}^{2} { \sin }^{2} \theta)(1 - {n}^{2} { \sin }^{2} \theta)(1 - {n}^{2} { \sin }^{2} \theta).....(iii)$ $(m' + m \cos \theta) (n' + n \cos \theta) = m'n' + (m'n + mn') \cos \theta + mn \: { \cos }^{2} \theta....(iv)$ $but \: mn + m'n' + (mn' + nm') \cos \theta = 1 \: in \: eq(iv)$ $(m' + m \cos \theta)(n'n \cos \theta) = 1 - mn + mn { \cos }^{2} \theta$ $= > 1 - mn(1 - { \cos }^{2} \theta)$ $= > 1 - mn \: { \sin }^{2} \theta...(v)$ $eq(v {)}^{2} \: we \: get$ $(m' + m \cos \theta)(n' + n \cos \theta) = 1 - mn + mn \: { \sin}^{2} \theta {)}^{2} .....(vi)$ $from \: eq(iii) \: and \: (vi)$ ${m}^{2} {n}^{2} { \sin}^{2} \theta = (1 - {m}^{2} { \sin }^{2} \theta)(1 - {n}^{2} { \sin }^{2} \theta)$ $\cancel{{m}^{2} {n}^{2} { \sin }^{4} \theta }= 1 - {n}^{2} { \sin }^{2} \theta - {m}^{2} { \sin}^{2} \theta + \cancel{ {m}^{2} {n}^{2} { \sin }^{4} }$ ${m}^{2} { \sin }^{2} \theta + {m}^{2} { \sin }^{2} \theta = 1$ ${ \sin }^{2} \theta( {m}^{2} + {n}^{2} ) = 1$ ${m}^{2} + {n}^{2} = \frac{1}{ { \sin }^{2} \theta }$ $\boxed{ {m}^{2} + {n}^{2} = { \cosec }^{2} \theta}$ *HENCE* *PROVED*.........

Answer»

$\boxed {\mathbb{ \red{G} \blue{I} \pink{V} \purple{E} \orange{ N}}}$

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

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

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

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

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

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

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

$eq(i) \times (ii)$

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

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

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

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

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

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

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

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

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

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

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