Higher Engineering Mathematics, Sixth Edition

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.

1. sinxcotx=cosx

2.

1

√

( 1 −cos^2 θ)

=cosecθ