Question

7 cosecA -3 cotA =7 prove that 7 cotA - 3 cosecA =3

Answer 1

$\textbf{Given:}$

$7\,cosecA-3\,cotA=7$

$\text{Squaring on both sides, we get}$

$(7\,cosecA-3\,cotA)^2=49$

$\text{using}$

$\boxed{\bf(a-b)^2=a^2+b^2-2ab}$

$49\,cosec^2A+9\,cot^2A-42\,cosecA\,cotA=49$

$\text{using}$

$\boxed{\begin{minipage}{4cm} \bf\,cosec^2A=1+cot^2A\\\\cot^2A=cosec^2A-1 \end{minipage}}$

$49(1+cot^2A)+9(cosec^2A-1)-42\,cosecA\,cotA=49$

$49+49\,cot^2A+9\,cosec^2A-9-42\,cosecA\,cotA=49$

$49\,cot^2A+9\,cosec^2A-9-42\,cosecA\,cotA=0$

$\implies(7\,cot\,A)^2+(3\,cosecA)^2-9-2\,(7\,cot\,A)(3\,cosec\,A)=9$

$\implies(7\,cot\,A-3\,cosecA)^2=9$

$\text{Take square root on both sides we get}$

$\implies\boxed{\bf\,7\,cot\,A-3\,cosecA=3}$

Answer 2

Answer:

Hi your answer is as follows

Step-by-step explanation:

Given:

7 cosec A - 3cot A= 7

on squaring both sides we get,

$49 cosec^2 A + 9 cot^2 A+42cosecA*cotA=49\\as, cosec^2 - cot^2 = 1\\and , cosec^2 = 1+cot^2,cot^2 = 1+cosec^2\\49[1+cot^2 A] + 9[1+cosec^2 A]-42cosecA*cotA=49\\49 +49 cot^2 A + 9 +9cosec^2 A-42cosecA*cotA = 49\\49cot^2 A +9cosec^2 A -42cosecA*cotA=9 \\We~can~write~49cot^2A~as (7cotA)^2~and~9cosecA~as~(3cosecA)^2and~42cosec*cot~as~2(3cosecA)(7cotA)\\On~substituting~we~get,\\(7cotA)^2+(3cosecA)^2-2(3cosecA)(7cotA)=9\\As~it~is~in~the~form~of~(a-b)^2=a^2+b^2-2ab\\It~can~be~written~as,$

$(7cotA-3cosecA)^2=9$

on square rooting both sides we get,

7cotA - 3cosecA = 3

Hence proved

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