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

Solution:

Since A + B = 45o ∴ tan (A + B) = tan 45o = 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 - tanA) 

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 (60o – A) sin (60 + A)

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

= 1/2 sin A [cos 2 A – cos 120o]

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

Using 2 cos A sin B = sin (A + B) – sin (A – B) and cos 120o = -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 20o is equal to

(1) 2

(2) 2 sin 20°/ sin 40°

(1) 4

(2) 4 sin 20°/ sin 40°

Solution:

√3 cosec 20° - sec 20o

= tan 60° cosec 20° - sec 20o

= [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:


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.

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