Question 1 :
Show that
tan2θ - sin2θ = tan2θsin2θ
Solution :
tan2θ - sin2θ = tan2θ - sin2θ(cos2θ/cos2θ)
= tan2θ - (sin2θ/cos2θ)cos2θ
= tan2θ - tan2θcos2θ
= tan2θ(1 - cos2θ)
= tan2θsin2θ
Question 2 :
Show that
secθ - cosθ = tanθsinθ
Solution :
secθ - cosθ = 1/cosθ - cosθ
= (1 - cos2θ)/cosθ
= sin2θ/cosθ
= (sinθ/cosθ)sinθ
= tanθsinθ
Question 3 :
Show that
sin2x + cos2x - 1 = 2sinx(cosx - sinx)
Answer :
sin2x + cos2x - 1 :
= 2sinxcosx + cos2x - sin2x - sin2x - cos2x
= 2sinxcosx - sin2x - sin2x
= 2sinxcosx - 2sin2x
= 2sinx(cosx - sinx)
Question 4 :
Show that
cos4θ - sin4θ = cos2θ
Solution :
cos4θ - sin4θ = (cos2θ)2 - (sin2θ)2
Using the identity a2 - b2 = (a + b)(a - b),
= (cos2θ + sin2θ)(cos2θ - sin2θ)
= (1)(cos2θ)
= cos2θ
Question 5 :
Show that
sin4θ + cos4θ = 1 - 2sin2θcos2θ
Solution :
sin4θ + cos4θ = (sin2θ)2 + (cos2θ)2
Using the identity a2 + b2 = (a + b)2 - 2ab,
= (sin2θ + cos2θ)2 - 2sin2θcos2θ
= 12 - 2sin2θcos2θ
= 1 - 2sin2θcos2θ
Question 6 :
Show that the equation 2cos2x + 5sinx = 4 may be written in the form
2sin2x - 5sinx + 2 = 0
Answer :
2cos2x + 5sinx = 4
2(1 - sin2x) + 5sinx = 4
2 - 2sin2x + 5sinx = 4
-2sin2x + 5sinx + 2 = 4
Multiply both sides by -1.
-1(-2sin2x + 5sinx + 2) = -1(4)
2sin2x - 5sinx - 2 = -4
Add 4 to both sides.
2sin2x - 5sinx + 2 = 0
Question 7 :
Show that
Answer :
Question 8 :
Show that
Answer :
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