Question

I want the prove please help me​

Answer 1

hope its help.........

Answer 2

$\huge{ \underline{ \underline{ \green{ \sf{ Detailed \: Answer :}}}}}$

• To prove :

$\frac{ \tan {}^{3} (x) }{1 + \tan {}^{2} (x) } + \frac{ \cot {}^{3} (x) }{1 + \cot {}^{2} (x) } = \sec(x) \cosec(x) - 2 \sin( x ) \cos(x)$

proof :

LHS :

$\frac{ \tan {}^{3} (x) }{1 + \tan {}^{2} (x) } + \frac{ \cot {}^{3} (x) }{1 + \cot {}^{2} (x) }$

$= \frac{ \frac{ \sin {}^{3} (x) }{ \cos {}^{3} (x) } }{1 + \frac{ \sin {}^{2} (x) }{ \cos {}^{2} (x) } } + \frac{ \frac{ \cos {}^{3} (x) }{ \sin {}^{3} (x) } }{1 + \frac{ \cos {}^{2} ( x ) }{ \sin {}^{2} ( x ) } }$

$= \frac{ \frac{ \sin {}^{3} (x) }{ \cos {}^{ 3} (x) } \times \cos {}^{2} (x) }{ \cos {}^{2} (x) + \sin {}^{2} (x) } + \frac{ \frac{ \cos {}^{3} (x) }{ \sin {}^{3} (x) } \times \sin {}^{2} (x) }{ \cos {}^{2} (x) + \sin {}^{2} (x) }$

since , sin ²x + cos²x = 1

$= \frac{ \sin {}^{3} (x) }{ \cos(x) } + \frac{ \cos {}^{3} (x) }{ \sin {}^{} (x) }$

$= \frac{ \sin {}^{4} (x) + \cos {}^{4} (x) }{ \cos(x) \sin(x) }$

$= \frac{ (\sin {}^{2} (x) + \cos {}^{2} (x)) {}^{2} - 2 \sin {}^{2} (x) \cos {}^{2} (x) }{ \cos(x) \sin(x) }$

$= \frac{1 - 2 \sin {}^{2} (x) \cos {}^{2} (x) }{ \sin(x) \cos(x) }$

$= \frac{1}{ \sin(x) \cos(x) } - \frac{2 \sin {}^{2} (x) \cos {}^{2} (x) }{ \sin( x ) \cos(x) }$

$= \sec(x) \csc(x) - 2 \sin(x) \cos(x)$

$= RHS$

$\huge{\bold{ Hence proved }}$