F.2 maths x2

1. √(1 + tan^2 θ) = 1 / cos θ

左方:
√( 1 + tan^2 θ)


=√ (1 + sin^2 θ over cos^2 θ )

=√(cos^2 θ over cos^2 θ + sin^2 θ over cos^2 θ )

=√( 1 over cos^2 )

=1 over cos θ

右方:

= 1 over cos θ

所以 right = leaf

1. √(1 + tan^2 θ) = 1 / cos θ

√(1+tan^2 θ)
= √(1+sin^2 θ/cos^2 θ)
= √[(cos^2 θ + sin^2 θ)/cos^2 θ]
= √(1/cos^2 θ)
= 1/cos θ

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

2. 5+ 4sin(90 º - θ) – sin^2 θ = (cosθ + 2)^2

5+ 4sin(90 º - θ) – sin^2 θ
= (4+1) + 4cos θ - sin^2 θ
= (4+ cos^2 θ + sin^2 θ) + 4cos θ - sin^2 θ
= 4 + cos^2 θ + 4cos θ
= (cosθ + 2)^2

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Note that:
1. sin^2 θ + cos^2 θ = 1
2. sin(90 º - θ) = cos θ

1.L.H.S=√(1 + tan^2 θ)
=√ (1+sin^2 θ/cos^2 θ)
=√ (cos^2 θ/cos^2 θ+ sin^2 θ/cos^2 θ)
=√ (cos^2 θ+sin^2 θ)/cos^2 θ
=√ (1/ cos^2 θ)
=1 / cos θ
=R.H.S
i.e. The proof is done.

2.L.H.S=5+ 4sin(90 º - θ) – sin^2 θ
=5+ 4 cos θ - (1- cos^2 θ)
=5+ 4 cos θ - 1 + cos^2 θ
=4+ 4 cos θ + cos^2 θ
=(cos θ + 2 )^2 <- - - - -(By using Fator method)
=R.H.S
i.e. The proof is done.

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1.LHS=√(1+sin^2θ/cos^2θ)
.........=√((sin^2θ+cos^2θ)/cos^2θ)
.........=√(1/cos^2θ)
.........= 1 / cos θ
.........=RHS
2.LHS=5+ 4sin(90 º - θ) – sin^2 θ
.........=5-(1-cos^2θ)+4cosθ
.........=cos^2θ+4cosθ+4.....................let x = cosθ,,x^2+4x+4.......(x+2)^2
.........=(cosθ + 2)^2
.........=RHS

1. √(1 + tan^2 θ)
= √(1+ sin^2 θ/cos^2 θ)
= √[(cos^2 θ + sine^2 θ)/cos^2 θ]
= √(1/cos^2 θ)
= √[(1/cos θ)^2]
= 1/cos θ

2. 5+ 4sin(90 º - θ) – sin^2 θ
= 5 + 4cosθ - (1-cos^2 θ)
= 5 + 4cosθ - 1 + cos^2 θ
= 4 + 4cosθ + cos^2 θ
= (cos θ + 2)^2