# If cosecA-sinA=a^3If secA-csoA=b^3What is a^2xb^2(a^2+b^2)

Sudheesh Singanamalla
114 Points
13 years ago

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

Pushkar Pandit
31 Points
13 years ago

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