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Question

If x=a sec A cos B, y=b sec A sin B and z=c tan A, then x2a2+y2b2z2c2=


A

0

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B

3

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C

2

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D

1

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Solution

The correct option is D

1


x=a sec A cos Bxa=sec A cos B

y=b sec A sin Byb=sec A sin B

z=c tan Azc=tan A

Squaring each of the equations we get,

x2a2=sec2 A cos2 B, y2b2=sec2 A sin2 B,z2c2=tan2 A

x2a2+y2b2z2c2=sec2 A cos2 B+sec2 A sin2 Btan2 A

=sec2 A(cos2 B+sin2 B)tan2 A

=sec2 Atan2 A(cos2 B+sin2 B=1)

=1(sec2 Atan2 A=1)


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