Question

sin^23A/sin^2A-cos^23A/cos^2A=8cos^2A

Answer 1

Qᴜᴇsᴛɪᴏɴ :-

Prove :- (sin²3A/sin²A) - (cos²3A/cos²A) = 8cos2A

Sᴏʟᴜᴛɪᴏɴ :-

Solving LHS,

→ (sin²3A/sin²A) - (cos²3A/cos²A)

Taking LCM,

→ (sin²3A*cos²A - cos²3A*sin²A) /(sin²A * cos²A)

→ {(sin²3A*cos²A) - (cos²3A*sin²A)} /(sin²A * cos²A)

→ {(sin3A*cosA)² - (cos3A*sinA)²} /(sin²A * cos²A)

using (a + b)(a - b) = a² - b² in Numerator,

→ {(sin3A*cosA + cos3A*sinA)(sin3A*cosA - cos3A*sinA)} /(sin²A * cos²A)

using

• sin(A + B) = sinAcosB + cosAsinB
• sin(A - B) = sinAcosB - cosAsinB

in Numerator,

→ {sin(3A + A)*sin(3A - A)} /(sin²A * cos²A)

→ (sin4A*sin2A) /(sin²A * cos²A)

Multiply and divide by 4 on Both Numerator & Denominator Now,

→ 4(sin4A*sin2A) /4(sin²A * cos²A)

→ 4(sin4A*sin2A) /(2sinA*cosA)²

using 2sinA*cosA = sin2A in Denominator Now,

→ 4(sin4A*sin2A) /(sin2A)²

→ 4(sin4A*sin2A) /(sin²2A)

Again using 2sinA*cosA = sin2A in Numinator Now,

→ 4sin2A(2sin2A*cos2A) /(sin²2A)

→ (8*sin²2A * cos2A) /(sin²2A)

→ 8cos2A = RHS (Proved).

Answer 2

${ \rm{ \huge{ \underline{ \underline{ \green{QUESTION: -}}}}}}$




To prove:-



${ \tt{ \red{ \frac{ { \sin(3A ) }^{2} }{ { \sin(A ) }^{2} } - \frac{ { \cos(3A) }^{2} }{ { \cos(A) }^{2} } = 8 { \cos(A) }^{2} }}}$




${ \huge{ \rm{ \underline{ \underline{ \green{SOLUTION: -}}}}}}$




${ \tt{ \blue{ \frac{ { \sin(3A ) }^{2} }{ { \sin(A) }^{2} } - \frac{ { \cos(3A ) }^{2} }{ { \cos(A) }^{2} } = 8 { \cos(A) }^{2} }}} \\ \\ \\ { \sf{ \rightarrow{solving \: the \: L.H.S.}}}$




${ \red{ \dashrightarrow{ \tt{ \blue{ \frac{ { \sin(3A ) }^{2} }{ { \sin(A ) }^{2} } - \frac{ { \cos(3A ) }^{2} }{ { \cos(A) }^{2} } }}}}} \\ \\ \\ { \sf{ \rightarrow{taking \: LCM }}}$




${ \red{ \dashrightarrow{ \sf{ \blue{ \frac{( { \sin(3A) }^{2} \times { \cos(A) }^{2}) - ( { \cos(3A ) }^{2} \times { \sin(A ) }^{2} ) }{ { \sin(A) }^{2} \times { \cos(A) }^{2} } }}}}} \\ \\ \\ \\ { \red{ \dashrightarrow{ \sf{ \blue{ \frac{ {( \sin(3A ) \times \cos(A ) })^{2} - {( \cos(3A ) \times \sin(A) )}^{2} }{ { \sin(A) }^{2} \times { \cos(A) }^{2} } }}}}}$




${ \sf{ \rightarrow{using \: this\: algebraic \: identity}}} \\ \\ { \boxed{ \boxed{ \pink{ \sf{ {a}^{2} - {b}^{2} = (a + b)(a - b)}}}}} \\ \\ { \sf{ \: \: \: \: in \: the \: numerator}}$




${ \sf{ \blue{ \frac{{ \red{(}}({ \orange{ \sin(3A ) \times \cos(A)}}) + ({ \orange{ \cos(3A ) \times \sin(A) }}){ \red{)}}{ \red{(}}({ \orange{ \sin(3A ) \times \cos(A) }}) - ({ \orange{ \cos(3A ) \times \sin(A) }}){ \red{)}} }{ { \sin(A) }^{2} \times { \cos(A) }^{2} } }}}$




${ \sf{ \rightarrow{using \: this \: trignometric \: identity}}} \\ \\ { \star{ \underline{ \pink{ \sf{ \: \: \sin(A ) \cos(B ) + \cos(A ) \sin(B) = \sin(A + B ) }}}}} \\ \\ { \star{ \underline{ \pink{ \sf{ \: \: \sin(A) \cos(B) - \cos(A) \sin(B) = \sin(A - B ) }}}}} \\ \\ { \sf{ \: \: \: in \: the \: numerator}}$




${ \red{ \dashrightarrow{ \sf{ \blue{ \frac{( \sin(3A + A) )\times ( \sin(3A - A )) }{( { \sin(A) }^{2} \times { \cos(A ) }^{2} } }}}}} \\ \\ \\ { \red{ \dashrightarrow{ \blue{ \sf{ \frac{ \sin(4A) \times \sin(2A) }{ { \sin(A) }^{2} \times { \cos(A) }^{2} } }}}}}$




${ \sf{ \rightarrow{on \: multiplying \: and \: dividing \: by \: 4}}} \\ { \sf{ \: \: to \: both \:}} \\ { \sf{ \: \: numerator \: and \: denominator}} \\ { \sf{ \: \: we \: get....}}$




${ \red{ \dashrightarrow{ \sf{ \blue{ \frac{4 \times \sin(4A ) \times \sin(2A ) }{4( { \sin(A) }^{2} \times { \cos(A) }^{2})}}}} } } \\ \\ \\ \\ { \red{ \dashrightarrow{ \blue{ \sf{ \frac{4 \times \sin(4A ) \times \sin(2A ) }{ {(2 \times \sin(A ) \times \cos(A) ) }^{2} } }}}}}$




${ \sf{ \rightarrow{using \: the \: trignometric \: identity}}} \\ \\ { \star{ \underline{ \pink{ \sf{ \: \: 2 \sin(A ) \times \cos(A) = \sin(2A) }}}}} \\ \\ { \sf{ \: \: in \: the \: denominator}}$




${ \red{ \dashrightarrow{ \blue{ \sf{ \frac{4 \times (\sin(4A ) \times \sin(2A )) }{ {( \sin(2A ) )}^{2} } }}}}}$




${ \sf{ \rightarrow{now \: using}}} \\ \\ { \star{ \underline{ \pink{ \sf{ \: \: \: 2 \sin(A ) \cos(A) = \sin(2A) }}}}} \\ \\ { \sf{ \: \: in \: the \: numerator}}$




${ \red{ \dashrightarrow{ \blue{ \sf{ \frac{4 \sin(2A )(2 \sin(2A ) \times \cos(2A ) )}{ { \sin(2A) }^{2} } }}}}} \\ \\ \\ { \red{ \dashrightarrow{ \blue{ \sf{ \frac{8 { \sin(2A ) }^{2} \times \cos(2A ) }{ { \sin(2A ) }^{2} }}}}}}$





${ \red{ \dashrightarrow{ \blue{ \sf{8 \cos(2A ) }}}}}$




${ \sf{therefore}} \\ \\ { \boxed{ \red{ \sf{L.H.S. = R.H.S. }}}}$




${ \boxed{ \boxed{ \red{ \sf{ \frac{ { \sin(3A) }^{2} }{ { \sin(A) }^{2} } - \frac{ { \cos(3A) }^{2} }{ { \cos(A) }^{2} } = 8 \cos(2A ) }}}}} \\ \\ \\ { \sf{ \rightarrow{ \green{hence. \: proved}}}}$