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Vishnu Saran Grade: 10
        

If cosecA-sinA=a^3


If secA-csoA=b^3


What is a^2xb^2(a^2+b^2)


 

7 years ago

Answers : (2)

Sudheesh Singanamalla
114 Points
										

Dear Vishnu ,

a^3 = cosecA - sinA

b^3 = secA - cosA

to find : a^2 * b^2 ( a^2 + b^2)

proof :

a^2 = cosec A - sin A / a ;

b^2 = secA - cos A / b ;

substituting in a^2 * b^2 (a^2 + b^2) we get

cosec A - sin A / a * sec A - cos A/b [ cosec A - sin A / a + sec A - cos A /b ]

cos^2 A/ a sin A * sin^2 A / b cos A [ cos^2 A / a sin A + sin^2 A / b cos A ]

sin A . cos A / ab * [ bcos^3 A + a sin^3 A / a*b*sin A* cos A ]

b cos^3 A + a sin^3 A / a^2 * b^2

 

this is the answer !

 

Please approve if the answer is correct

7 years ago
Pushkar Pandit
31 Points
										

a^3=cosecA-sinA =>a^2=cosecA-sinA/a

b^3=secA-cosA  =>b^2=secA-cosA/b

Now,

a^2=(1/sinA-sinA)/a

     =(1-sin^2 A/sinA)/a

     =(cos^2 A/sinA)/a

     =cotA*cosA/a

b^2=secA-cosA/b

     =(1/cosA-cosA)/b

     =(1-cos^2 A/cosA)/b

     =(sin^2 A/cosA)/b

     =tanA*sinA/b

We have to find the value of 

a^2xb^2(a^2+b^2)

So lets solve it

=>cotA*cosA/a . tanA*sinA/b{(cotA*cosA/a)+(tanA*sinA/b)}

=>cosA*sinA/ab . {cos^2 A/sinAa +sin^2 A/cosAb  }

=>cosA*sinA/ab . {cos^2 A+sin^2 A/sinA.cosA.ab}

=>cosA*sinA/ab . {1/sinAcosAab}      [sin^2 A + cos^2 A = 1]

=>1/(ab)2



6 years ago
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