Page 36 - mathsvol1ch1to3ans

36

2 sin α 1 − cos α + sin α
6. If y = , then prove that = y.
1 + cos α + sin α 1 + sin α

Solution:

α
1 − cos α + sin α 2 sin 2 α + 2 sin cos α 2 sin α
= 2 2 2 = 2
α 2
1 + sin α (sin α + cos ) sin α + cos α
2 2 2 2
α
2 sin α 4 sin cos α 2 sin α
y = = 2 2 = 2
α
1 + cos α + sin α 2 cos 2 α + 2 sin cos α sin α + cos α
2 2 2 2 2
P ∞ 2n P ∞ 2n P ∞ 2n 2n π
7. If x = cos θ, y = sin θ and z = cos θ sin θ, 0 < θ < , then show that
n=0 n=0 n=0 2
1
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3
2
xyz = x + y + z. [Hint: Use the formula 1 + x + x + x + . . . = , where |x| < 1].
1 − x
Solution:
1
2
3
Using the formula 1 + x + x + x + . . . = we get,
1 − x
1 1 1
x = , y = and z = .
2
2
2
2
1 − cos θ 1 − sin θ 1 − sin θ cos θ
1 1 1 1 1 1 1 1 1
+ + = + +
yz xz xy y z x z x y
= (1 − sin θ) (1 − sin θ cos θ)
+ (1 − cos θ) (1 − sin θ cos θ) + (1 − cos θ) (1 − sin θ) ********************************************************************
2
= 1 − sin θ − sin θ cos θ + sin θ cos θ
2
+1 − cos θ − sin θ cos θ + cos θ sin θ + 1 − cos θ − sin θ + sin θ cos θ
*******************************************************************
********************************************************************

2
3
2
8. If tan θ = 1 − k , show that sec θ + tan θ cosec θ = 2 − k 2 3/2 . Also, find the values of k for
which this result holds.
2
Solution:Given tan θ = 1 − k 2
2
3
2
2
1 sin θ 1 1 sin θ cos θ + sin θ 1
3
sec θ + tan θcosec θ = + × = + = =
3
3
3
3
cos θ cos θ sin θ cos θ cos θ cos θ cos θ
3
3 3
2
2
3
2
= sec θ = (sec θ) 2 = (1 + tan θ) 2 = (2 − k )2
9. If sec θ + tan θ = p, obtain the values of sec θ, tan θ and sin θ in terms of p.
Solution:
1
sec θ − tan θ =
p
2
p + 1
sec θ =
2p
2
p − 1
tan θ =
2p
2
tan θ p − 1
sin θ = =
2
sec θ p + 1