Solved Examples on Trigonometric Ratios

Trigonometry is basically the study of angles. Various double and half angle formulas as well as multiple angle formulas are extremely important as they often fetch direct questions in various exams like the IIT JEE. Some of the illustrations based on these formulae are listed below:

Illustration:

If A + B = 45^o, Then show that (1 + tan A) (1 + tan B) = 2

Solution:

Since A + B = 45^o ∴ tan (A + B) = tan 45^o = 1

or, (tan A + tan B) / (1 - tan A tan B)  = 1

This gives, tan A + tan B = 1 – tan A tan B

So, tan A + tan B + tan A tan B = 1

tan A  (1 + tan B) + tan B + 1 = 2

Hence, (1 + tan B) (1 + tan A) = 2 

Some Results obtained from compound formulae:

Result (i) sin (A + B) sin (A – B) = sin² A – sin² B = cos²B – cos² A

Proof: First we consider the L.H.S 

L.H.S = (sin A cos B + cos A sin B) (sin A cos B – cos A sin B)

= sin²A cos²B – cos²A sin²B

= sin²A (1 – sin²B) – (1 – sin²A) (sin²B)

= sin²A – sin²A sin²B – sin²B + sin²A sin²B

= sin²A – sin²B

= RHS

= 1 – cos² A – 1 + cos² B

= cos²B – cos²A.

Result (ii) cos (A + B) cos (A – B) = cos² A – sin² B

Proof:  (cos A cos B – sin A sin B) (cos A cos B + sin A sin B)

= cos² A cos² B – sin² A sin² B

= cos²A (1 – sin²B) – (1 – cos²A) sin²B

= cos²A – cos²A sin²B – sin²B + sin²B cos²A

= cos2A – sin2B

 Result (iii)

(a) sin 2 A = 2tan A / (1+ tan² A)  (Dividing and multiplying by cos² A)

(b) cos 2 A = (1 - tan² A) / (1 + tan² A)

(c) tan 2 A = 2 tan A / (1 - tan² A) 

Illustration:

If 3 tan θ tan ∅ = 1, then prove that 2 cos (θ + ∅) = cos (θ – ∅)

Solution:

It is given that 3 tan θ tan ∅ = 1

or,  cot θ cot ∅ = 1/3

or,  cos θ cos ∅ / sin θ sin ∅ = 1/3      

Using componendo and dividendo, we have.

(cos θ cos ∅ + sin θ sin ∅) / (cos θ cos ∅ - sin θ sin ∅) = 4/2                      

This gives cos (θ - ∅)/ cos (θ + ∅) = 2

or,  2 cos (θ + ∅) = cos (θ – ∅).

Illustration:

Prove that sin (60o – A) sin A sin (60o + A) = 1/4 sin 3 A

Solution:

L.H.S. = sin A sin (60^o – A) sin (60 + A)

= 1/2 sin A [2 sin (60 – A) sin (60 + A)]

= 1/2 sin A [cos 2 A – cos 120^o]

= 1/4 [2 sin A cos 2 A – 2 cos 120^o sin A]

Using 2 cos A sin B = sin (A + B) – sin (A – B) and cos 120^o = -1/2.

We get,

= 1/4 [sin (2A + A) – sin (2A – A) – 2 (- 1/2) sin A]

= 1/4 sin 3A

= R.H.S

Illustration:

The value of the expression √3 cosec 20° - sec 20^o is equal to

(1) 2

(2) 2 sin 20°/ sin 40°

(1) 4

(2) 4 sin 20°/ sin 40°

Solution:

√3 cosec 20° - sec 20^o

= tan 60° cosec 20° - sec 20^o

= [sin 60° cos 20° - cos 60° sin 20°] / cos 60° sin 20° cos 20°

= sin (60° - 20°) / cos 60° sin 20° cos 20°

= sin 40° / sin 20° cos 20° (1/2)

= 2 sin 20° cos 20° / sin 20° cos 20° (1/2)

= 4.

Illustration:

If tan A = (1 - cos B) / sin B, then is tan 2A = tan B?

Solution:

Given that tan A = (1 - cos B) / sin B

                              = (2 sin² B/2)/ 2 sin B/2 cos B/2

 Hence, this gives tan A = tan B/2

Hence, this implies tan 2A= tan B.

Hence, the given statement is correct.

Students are advised to memorize the entire double angle and half angle formulas as majority of the questions asked in the various examinations like the IIT JEE are based on the direct or indirect application of these formulas. It is important to know all the trigonometry formulas in order to remain competitive in the JEE. 

Related Resources:

• Look into the Revision Notes on Trigonometry for a quick revision.
• Various Recommended Books of Mathematics are just a click away.

To read more, Buy study materials of Inverse Trigonometric Functions comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics here.