Question

(cot A + 3 ) (3 cot A + 1 ) = 3 cosec²A + 10 cot A

Is this statement correct​

Answer 1

Answer:

Yes, the statement is correct.

Step-by-step explanation:

$( \frac{ \cos(a) }{ \sin(a) } + 3)( 3 + \frac{ \cos(a) }{ \sin(a) } + 1) = 3( \frac{1}{ \sin(a) } )^{2} + 10 \frac{ \cos(a) }{ \sin(a) }$

$( \frac{ \cos(a) + 3 \sin(a) \times 3 \cos(a) + \sin(a) }{ \sin(a)^{2} } ) = 3( \frac{1}{ \sin(a) } )^{2} + 10 \frac{ \cos(a) }{ \sin(a) }$

1

Remove parenthesis

$( \frac{3 \cos(a)^{2} + \sin(a) \cos(a) + 9 \sin(a) }{ \cos(a) + 3 \sin(a)^{2} } = 3( \frac{1}{ \sin(a) } )^{2} + 10 \frac{ \cos(a) }{ \sin(a) }$

$\frac{3 \cos(a) \sin(a) + 9 \cos(a) \sin(a) }{ \sin(a)^{2} } = 3( \frac{1}{ \sin(a) } )^{2} + 10 \frac{ \cos(a) }{ \sin(a) }$

$\frac{3 + 10 \cos(a) \sin(a) }{ \sin( {a}^{2} ) } = 3( \frac{1}{ \sin(a) } )^{2} + 10 \frac{ \cos(a) }{ \sin(a) }$

$\frac{3 + 5 \sin(2a) }{ \sin( {a}^{2} ) } - ( \frac{3}{ \sin(a) } )^{2} - \frac{ 10\cos(a) }{ \sin(a) } = 0$

$\frac{3 + 5 \sin(2a) - 3 - 10 \sin(a \cos(a) ) }{ \sin{(a)}^{2} } = 0$

${3 + 5 \sin(2a) - 3 - 10 \sin(a \cos(a) ) = 0$

$5 \sin(2a) - 5 \sin(2a) = 0$

$0 = 0$