Question

If cot theta+cos theta =m and cot theta - cos theta =n then prove tht (m^2-n^2)^2=16mn

Answer 1

HELLO D,


cottheta + costheta = m

cottheta - costheta = n

now,


(m² - n²)² = [(m - n)(m + n)]²

[ (cottheta + costhet - cottheta + costheta)(cottheta - costhet + cottheta + costheta) ]²

[ (2costheta)(2cottheta) ] ²

[ 16(cot²theat) . (cos²theta) ]-----------( 1 )




now,


16mn

16(cottheta + costheta)(cottheta - costheta)

16(cot²theta - cos²theta)

16(cos²theta / sin²theta - cos²theta)

16(cos²theta - cos²theta . sin²theta)/sin²theta

16(cos²theta(1 - sin²theta)/sin²theta

16(cos²theta . cos²theta)/sin²theta

16(cos²theta/sin²theta . cos²theta)

16(cot²theta . cos²theta)------------( 2 )




from-----( 1 ) & -----( 2 )


[(m² - n²)]² = 16(mn)





I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Answer: note :@=theta

Firstly put the value of mn in ( m^2-n^2)^2

We get,

={(Cot@+cos@)^2-(cot@-cos@)^2}^2

we know that,a^2-b^2=(a+b)(a-b)

={(Cot@+cos@)-(cotx+cos@)-(cot@-cos@)(cot@+cos@)}

={(2cos@)(2cot@)}

={16(cot^2@)×(cos^2@)}. -----

(1st equation)

=16(cot@+cos@)(cot@-cos@)

=16(cot^2@-cos^2@)

=16(cos^2@/sin^2-cos^2@/1)

=16(cos^2@-cos^2×sin^2@)

=16{cos^2@(1-sin^2)/sin^2@}

=16(cos^2@×cos^2@/sin^2@)

=16(cos^2@/sin^2@×cos^2@

=16(cot^2@×cos^2@)-------(2) equation

From (1) and (2)

[(m^2-n^2)]^2=16mn

Step-by-step explanation: