Higher Engineering Mathematics

TRIGONOMETRIC IDENTITIES AND EQUATIONS 167

B

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 −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 −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=1 then 1−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 73 Further problems on trigono-

metric identities

In Problems 1 to 6 prove the trigonometric

identities.

1. sinxcotx=cosx

2.

1

√

(1−cos^2 θ)

=cosecθ

3. 2 cos^2 A− 1 =cos^2 A−sin^2 A

4.

cosx−cos^3 x

sinx

=sinxcosx

5. (1+cotθ)^2 +(1−cotθ)^2 =2 cosec^2 θ

6.

sin^2 x( secx+cosecx)

cosxtanx

= 1 +tanx

16.3 Trigonometric equations

Equations which contain trigonometric ratios are

calledtrigonometric equations. There are usually

an infinite number of solutions to such equations;

however, solutions are often restricted to those

between 0◦and 360◦.

A knowledge of angles of any magnitude is essen-

tial in the solution of trigonometric equations and

calculators cannot be relied upon to give all the