sinA*cosA(tanA+cotA)=1
Answers
Let ;-
A = ( 1 , -1 , 3)
B = ( 2 , -4 , 5)
C = ( 5 , -13, 11)
If , all these points lie on the straight lines than ;-
=> AC = AB + BC
=> AB = √( 2-1 )² + ( -4 + 1)² +( 5-3 )²
=> AB = √ 1 + 9 + 4
=> AB = √ 14
=> AB = 2√5
=> BC = √ ( 5 - 2 )² + ( - 13 + 4 )² + ( 11 - 5 )²
=> BC = √ 9 + 81 + 36 = 3 √ 14
=> AC = √ ( 5 - 1 )² + ( - 13 + 1 )² + ( 11 - 3 )²
=> AC = √ 16 + 144 + 64
=> AC = 4 √ 14
=> AB + BC = AC
=> L. H. S = √ 14 + 3√ 14
=> L. H. S. = 4 √ 14 = AC
=> L. H. S. = R. H. S.
Therefore,
The given points are in a Straight line.
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Correct option is
B
secA+cosecA
The value of (sinA+cosA)(tanA+cotA) is
=(sinA+cosA)(
cosA
sinA
+
sinA
cosA
)
=(sinA+cosA)(
cosAsinA
sin
2
A+cos
2
A
)
=
cosAsinA
sinA+cosA
=
cosAsinA
sinA
+
cosAsinA
cosA
=
cosA
1
+
sinA
1
=secA+cosecA
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