Trigonometric Ratios & Identities- 2 | Mathematics (Maths) for JEE Main & Advanced PDF Download

Trigonometric Ratios of the Sum & Difference of Two Angles

➢ Some More Results

Solved Examples

Example.5. Prove that √3 cosec20° – sec20° = 4.
Solution. L.H.S.




Example.6. Prove that tan70° = cot70° + 2cot40°.
Solution. L.H.S.

= tan70° = tan (20° + 50°)

or tan70° – tan20° tan50° tan70° = tan20° + tan50°
or tan70° = tan70° tan50° tan20° + tan20° + tan50° = 2 tan 50° + tan20°
= cot70° + 2cot40° = R.H.S.

Formulas to Transform the Product into Sum or Difference

Example.7. If sin2A = λ sin2B, then prove that 
Solution. Given  sin2A = λ sin2B

Applying componendo & dividendo,

Formulas to Transform Sum or Difference into Product

Example.8.is equal to:

Trigonometric Ratios of sum of more than two angles

(i) sin (A + B + C) = sinAcosBcosC + sinBcosAcosC + sinCcosAcosB – sinAsinBsinC
= ΣsinA cosB cosC – Πsin A
 cosA cosB cosC [tanA + tanB + tanC - tanA tanB tanC]
(ii) cos (A + B + C) = cosA cosB cosC – sinA sinB cosC – sinA cosB sinC – cosA sinB sinC
= Πcos A  - Σsin A sin B cos C
= cos A cos B cos C [1 - tan A tan B - tan B tan C - tan C tan A]
(iii) tan (A + B + C) =

Trigonometric Ratios of Multiple Angles

(a) Trigonometrical ratios of an angle 2θ in terms of the angle θ :
(i) sin 2θ = 2 sin θ cos θ =
(ii) cos 2θ = cos²θ - sin²θ = 2 cos²θ - 1 = 1 - 2 sin²θ =
(iii) 1 + cos 2θ = 2 cos² θ
(iv) 1 - cos 2θ = 2 sin² θ
(v)
(vi)

Example.10. Prove that := tan (60º + A) tan (60º - A)
Solution. R.H.S. = tan(60° + A) tan(60° – A)

(b) Trigonometrical ratios of an angle 3θ in terms of the angle θ :
(i) sin3θ = 3sinθ - 4sin³θ.
(ii) cos3θ = 4cos³θ - 3cosθ.
(iii)

Example.11. Prove that : tanA + tan(60° + A) + tan(120° + A) = 3tan3A
Solution. L.H.S. = tanA + tan(60° + A) + tan(120° + A)
= tanA + tan(60° + A) + tan{180° –(60° – A)}
= tanA + tan(60° + A) – tan(60° – A) [∵ tan(180° - θ) = -tanθ]

Trigonometric Ratios of Sub Multiple Angles

Example.12.is equal to
(a)
(b)
(c)
(d)
Ans. (a)
Solution. 

Trigonometric Ratios of Some Standard Angles

Example.13. Evaluate sin78 ° – sin66 ° – sin42 ° + sin 6°.
Solution. The expression = (sin78° - sin42°) - (sin66° - sin6°) = 2cos(60°) sin(18°) - 2cos36° . sin30°

Conditional Trigonometric Identities

If A + B + C = 180°, then
(i) tan A + tan B + tan C = tan A tan B tan C
(ii) cot A cot B + cot B cot C + cot C cot A = 1


(v) sin 2A + sin 2B + sin 2C = 4 sinA sinB sinC
(vi) cos 2A + cos 2B + cos 2C = –1 – 4 cosA cosB cosC



Example.14. In any triangle ABC, sin A – cos B = cos C, then angle B is
(a) π / 2
(b) π / 3
(c) π / 4
(d) π / 6
Ans. (a)
Solution. We have, sin A – cos B = cos C
sin A = cos B + cos C

  ∵ A + B + C = π


Therefore 2B = π ⇒ B = π / 2

Example.15. If A + B + C = 3π / 2, then cos 2A + cos 2B + cos2C is equal to
(a) 1 – 4cosA cosB cosC
(b) 4 sinA sin B sinC
(c) 1 + 2cosA cosB cosC
(d) 1 – 4 sinA sinB sinC
Ans. (d)
Solution. cos 2A + cos 2B + cos 2C  = 2 cos (A + B ) cos (A – B) + cos 2C
=cos (A – B) + cos 2C ∵ A + B + C = 3π / 2
= – 2 sin C cos ( A– B) + 1 –  2 sin²C  = 1 – 2 sinC [ cos ( A– B) + sin C)
= 1 – 2 sin C [ cos (A – B) + sin
= 1 – 2 sin C [ cos (A – B) – cos ( A +B ) ]  
= 1 – 4 sin A sin B sin C

Maximum & Minimum Values of Trigonometric Expressions

(i) acosθ + bsinθ will always lie in the interval  i.e. the maximum and minimum values are  respectively.
(ii) Minimum value of a² tan² θ + b² cot² θ = 2ab where a, b > 0
where α and β are known angles.(iv)and α + β = s (constant) then
(a) Maximum value of the expression cos α cos β, cos α + cos β, sin α sin β or sin α + sin β occurs when α = β = a / 2

(b) Minimum value of sec α + sec β, tanα + tanβ, cosec α + cosec β occurs when α = β = a / 2

(v) If A, B, C are the angles of a triangle then the maximum value of

sin A + sin B + sin C and sin A sin B sin C occurs when A = B = C = 60

(vi) In case a quadratic in sin θ & cos θ is given then the maximum or minimum values can be obtained by making a perfect square.

Example.16. Prove that: for all values of θ.
Solution. We have,


for all θ.
for all θ.
for all θ.

Example.17. Find the maximum value of
(a) 1
(b) 2
(c) 3
(d) 4
Ans. (d)
Solution. We have

∴ maximum value =

Important Results

(i) sinθ  sin (60° –  θ) sin (60° +  θ) =
(ii) cosθ. cos (60° -  θ) cos (60° + θ) =
(iii) tanθ tan (60° - θ) tan (60° + θ) = tan 3θ
(iv) cot θ cot (60° - θ) cot (60° + θ) = cot 3θ
(v) (a) sin² θ + sin² (60° +  θ) + sin² (60° -  θ) = 3 / 2
(b) cos² θ + cos² (60° +  θ) + cos² (60° - θ) = 3 / 2
(vi) (a) If tan A + tan B + tan C = tan A tan B tan C, then  A + B + C = nπ, n ∈ I
(b) If tan A tan B + tan B tan C + tan C tan A = 1, then  A + B + C = (2n + 1) π / 2, n ∈ I

(viii) (a) cotA – tanA = 2cot2A
(b) cotA + tanA = 2cosec2A



Example.18. Prove that tanA + 2tan2A + 4tan4A + 8cot8A = cot A.

Solution. 8 cot 8A = cotA - tanA - 2tan2A - 4tan4A
= 2 cot2A - 2tan2A - 4tan4A (using viii (a) in above results)
= 4 cot4A - 4tan4A (using viii (a) in above results)

= 8 cot8A.
_{Alternate Method:
L.H.S. = tanA + 2tan2A + 4tan4A +}




Example.19. Evaluate
 
Solution.


Example.20. Prove that : (1 + sec²θ)(1 + sec2²θ)(1 + sec2³θ)....(1 + sec2ⁿθ) = tan2ⁿθ.cotθ.
Solution. L.H.S.

Miscellaneous Illustration

Example.21. Prove that
tanα + 2 tan2α + 2² tan²α + . .. .. . + 2^{n - 1} tan 2^{n - 1} α  + 2ⁿ cot2ⁿα  = cot α

Solution. We know tanθ = cotθ  - 2 cot 2θ .....(i)

Putting 0 = a, 2α, 2²α,.....in (i), we get

tan α = (cot α - 2 cot 2α)

2 (tan 2 α) = 2(cot 2α - 2 cot 2²α)
22 (tan 2²α) = 22 (cot 2²α - 2 cot 2³α)
...........................................................
2^{n - 1} (tan 2^{n - 1}α) = 2^{n - 1} (cot 2^{n - 1}α - 2 cot 2ⁿα)

Adding,
tan α + 2 tan²α + 2² tan²α +....+ 2^n-1 tan 2^n-1α = cot a - 2ⁿ cot 2ⁿα
∴ tan α + 2 tan² α + 2² tan²α +....+ 2^n-1 tan 2^n-1 a + 2ⁿ cot 2ⁿ a = cot α

Example.22. If A,B,C and D are angles of a quadrilateral andprove that A = B = C = D = π / 2.
Solution. 


Since, A + B = 2π - (C + D), the above equation becomes,


This is a quadratic equation inwhich has real roots .





⇒ A = B , C = D .
Similarly A = C, B = D ⇒ A = B = C = D = π / 2