Trigonometric identities and equations 153
theright-handside(RHS)orvice-versa.It isoftenuseful
to change all of the trigonometric ratios into sines and
cosines where possible. Thus,
LHS=sin^2 θcotθsecθ
=sin^2 θ
(
cosθ
sinθ
)(
1
cosθ
)
=sinθ(by cancelling)=RHS
Problem 2. Prove that
tanx+secx
secx
(
1 +
tanx
secx
)= 1.
LHS=
tanx+secx
secx
(
1 +
tanx
secx
)
=
sinx
cosx
+
1
cosx
(
1
cosx
)
⎛
⎜
⎝^1 +
sinx
cosx
1
cosx
⎞
⎟
⎠
=
sinx+ 1
( cosx
1
cosx
)[
1 +
(
sinx
cosx
)(
cosx
1
)]
=
sinx+ 1
( cosx
1
cosx
)
[1+sinx]
=
(
sinx+ 1
cosx
)(
cosx
1 +sinx
)
=1(by cancelling)=RHS
Problem 3. Prove that
1 +cotθ
1 +tanθ
=cotθ.
LHS=
1 +cotθ
1 +tanθ
=
1 +
cosθ
sinθ
1 +
sinθ
cosθ
=
sinθ+cosθ
sinθ
cosθ+sinθ
cosθ
=
(
sinθ+cosθ
sinθ
)(
cosθ
cosθ+sinθ
)
=
cosθ
sinθ
=cotθ=RHS
Problem 4. Show that
cos^2 θ−sin^2 θ= 1 −2sin^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 −2sin^2 θ=RHS
Problem 5. Prove that
√(
1 −sinx
1 +sinx
)
=secx−tanx.
LHS=
√(
1 −sinx
1 +sinx
)
=
√{
( 1 −sinx)( 1 −sinx)
( 1 +sinx)( 1 −sinx)
}
=
√{
( 1 −sinx)^2
( 1 −sin^2 x)
}
Since cos^2 x+sin^2 x=1then1−sin^2 x=cos^2 x
LHS=
√{
( 1 −sinx)^2
( 1 −sin^2 x)
}
=
√{
( 1 −sinx)^2
cos^2 x
}
=
1 −sinx
cosx
=
1
cosx
−
sinx
cosx
=secx−tanx=RHS
Now try the following exercise
Exercise 65 Further problemson
trigonometric identities
In Problems 1 to 6 prove the trigonometric
identities.
- sinxcotx=cosx
2.
1
√
( 1 −cos^2 θ)
=cosecθ