Question

if tanA +sinA= m and tanA-sinA=n, then prove that (m²-n²)²=16nm

Answer 1

Please find in the attachment...

Hope it helps!!!

Answer 2

Step-by-step explanation:

Given tanA+sinA= m ---(1)

and tanA-sinA = n ---(2)

i) m+n = 2tanA ---(3)

ii)m-n=2sinA ----(4)

iii) m²-n²

=(m+n)(m-n) /* From (3)&(4)*/

= 2tanA × 2sinA

= 4tanAsinA ---(5)

Now,

LHS = (m²-n²)²

= (4tanAsinA)²

= 16tan²Asin²A

= 16tan²A(1-cos²A)

$=16[tan^{2}-tan^{2}Acos^{2}A}]$

$=16[tan^{2}A-\frac{sin^{2}A cos^{2}A}{cos^{2}A}]$

$=16[tan^{2}A-sin^{2}A}]$

$=16(tanA+sinA)(tanA-sinA)$

$=16mn\:[from\:(1)\:and\:(2) ]$

$=RHS$

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