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```        If cosecA-sinA=a^3
If secA-csoA=b^3
What is a^2xb^2(a^2+b^2)
```
7 years ago

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

```
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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