Trigonometric identities and equations 153
the right-hand side (RHS) or vice-versa. It is often useful
to change all of the trigonometric ratios into sines and
cosines where possible. Thus,
LHS = sin
2
θ cot θ sec θ
= sin
2
θ
cos θ
sin θ
1
cos θ
= sin θ (by cancelling) = RHS
Problem 2. Prove that
tan x + sec x
sec x
1 +
tan x
sec x
= 1.
LHS =
tan x + sec x
sec x
1 +
tan x
sec x
=
sin x
cos x
+
1
cos x
1
cos x
⎛
⎜
⎝1 +
sin x
cos x
1
cos x
⎞
⎟
⎠
=
sin x + 1
cos x
1
cos x
1 +
sin x
cos x
cos x
1
=
sin x + 1
cos x
1
cos x
[1 + sin x]
=
sin x + 1
cos x
cos x
1 + sin x
= 1 (by cancelling) = RHS
Problem 3. Prove that
1 + cot θ
1 + tan θ
= cot θ.
LHS =
1 + cot θ
1 + tan θ
=
1 +
cos θ
sin θ
1 +
sin θ
cos θ
=
sin θ + cos θ
sin θ
cos θ + sin θ
cos θ
=
sin θ + cos θ
sin θ
cos θ
cos θ + sin θ
=
cos θ
sin θ
= cot θ = RHS
Problem 4. Show that
cos 2 θ − sin 2 θ = 1 − 2 sin 2 θ.
From equation (2), cos 2 θ + sin 2 θ = 1, from which,
cos 2 θ = 1 − sin 2 θ.
Hence, LHS
= cos
2
θ − sin
2
θ = (1 − sin
2
θ) − sin
2
θ
= 1 − sin
2
θ − sin
2
θ = 1 − 2 sin
2
θ = RHS
Problem 5. Prove that
1 − sin x
1 + sin x
= sec x − tan x.
LHS =
1 − sin x
1 + sin x
=
(1 − sin x)(1 − sin x)
(1 + sin x)(1 − sin x)
=
(1 − sin x) 2
(1 − sin 2 x)
Since cos 2 x + sin
2 x = 1 then 1 − sin
2 x = cos 2 x
LHS =
(1 − sin x) 2
(1 − sin
2 x)
=
(1 − sin x) 2
cos 2 x
=
1 − sin x
cos x
=
1
cos x
−
sin x
cos x
= sec x − tan x = RHS
Now try the following exercise
Exercise 65 Further problems on
trigonometric identities
In Problems 1 to 6 prove the trigonometric
identities.
1. sin x cot x = cos x
2.
1
(1 − cos 2 θ)
= cosec θ