CBSE

1. TRIGONOMETRY
 Ratios of specific angles
 0/30/45/60/90

2. RATIOS FOR 45

3. RATIOS FOR 30 AND 60

4. TRIGONOMETRIC RATIOS OF SPECIFIC ANGLES
Angle A
00 300 450 600 900
Sin
0 ½ 1/√2 √3/2 1
Cos
1 √3/2 1/√2 1/2 0
Tan
0 1/√3 1 √3 undefined
Cot undefined √3 1 1/√3 0
Sec
1 2/√3 √2 2 undefined
Cosec undefined 2 1/√2 2/√3 1

6. Qn: If tan (A + B) = √3 and tan (A – B) =1/√3;
0° < A + B < 90°; A > B, find A and B.
Ans:
Given, tan (A + B) = √3 => tan(A+B) = tan60°
=> A + B = 60° => 1
tan (A- B) =1/√3 => tan(A- B) = tan30°
=> A - B = 30° => ②
1 + ②
A + B = 60°
A - B = 30°
2A = 90° => A = 90°/2 = 45°
from ② 45° + B = 60° => B = 60° - 45° = 15°
therefore, <A = 45° and <B = 15°

7. HOME WORK
Qn: 1.
If Sin(A + B) = 1 & Cos (A – B) =√3/2;
0° < A + B < 90°; A > B, find A and B.
Qn: 2.
If Cot(A + B) = 1/√3 & Cosec(A – B) = 2;
0° < A + B < 90°; A > B, find A and B.

8. TRIGONOMETRIC IDENTITIES

9. C In ∆ABC, ∟B = 90°
By Pythagorous theorem
B A AC2 = AB2 + BC2
=> AB2 + BC2 = AC2
AC2 AC2 AC2
Sin2A + Cos2A = 1
AC2 = AB2 + BC2
AB2 AB2 AB2
Sec2A = 1 + Tan2A
AC2 = AB2 + BC2
BC2 BC2 BC2
Cosec2A = Cot2A + 1

10. IDENTITY - 1
Sin2A + Cos2A = 1 => Sin2A = 1 - Cos2A
• => SinA = √ 1 - Cos2A
=> Cos2A = 1
- Sin2A
• => Cos A = √ 1 - Sin2A

11. IDENTITY - 2
 1 + Tan2A = Sec2A => Sec A = √ 1 + Tan2A
• => TanA = √ Sec2A - 1
 Sec2A - Tan2A = 1
 (SecA - TanA) (SecA + TanA) = 1
 (SecA - TanA) = 1/ (SecA + TanA)
 (SecA + TanA) = 1/ (SecA - TanA)

12. IDENTITY - 3
 1 + Cot2A = Cosec2A => Cosec A = √ 1 +
Cot2A
• => CotA = √Cosec2A -
1
 Cosec2A - Cot2A = 1
 (CosecA - CotA) (CosecA + CotA) = 1
 (CosecA - CotA) = 1/ (CosecA + CotA)
 (CosecA + CotA) = 1/ (CosecA - CotA)

13. QN 1. EXPRESS ALL THE TRIGONOMETRIC
RATIOS IN TERMS OF COTA
 Cot A
 Tan A = 1/Cot A
 CosecA = √ 1 – Cot2A
 Sin A = 1/ Cosec A = 1/√ 1 – Cot2A
 Cos A = √ 1 – Sin2A = √ 1 –( 1/√ 1 – Cot2A)2
= √1 –( 1/ 1 – Cot2A)=√(1– Cot2A –1)/ 1 – Cot2A)
= √ Cot2A/ (Cot2A-1) = CotA/√(Cot2A-1)
 Sec A = 1/CosA = 1/( CotA/√(Cot2A-1))
= √(Cot2A-1)/CotA

14. 5(v) Cos A – Sin A + 1 = Cosec A + Cot A
Cos A + Sin A – 1
L.H.S = Cos A – Sin A + 1
Cos A + Sin A – 1
dividing the Nr & Dr by Sin A, we get
= Cos A/Sin A – Sin A/Sin A + 1/Sin A
Cos A/Sin A + Sin A/Sin A – 1/Sin A
= CotA – 1 + Cosec A
CotA + 1 – Cosec A
= (CotA +Cosec A) - (Cosec A+CotA )( Cosec A-CotA)
CotA + 1 – Cosec A
= (CotA +Cosec A)[ 1 - Cosec A + CotA]
[ 1 - Cosec A + CotA]
= Cosec A + Cot A = R.H.S
Hence it is proved.

15. Prove that :
SinA – Cos A + 1 = Sec A + Tan A
Sin A + Cos A – 1

16. VII) Sinθ - 2Sin3 θ = Tanθ
2C0s3 θ - Cos θ
LHS = Sinθ - 2Sin3 θ
2C0s3 θ - Cos θ
= Sinθ - 2Sin3 θ
2C0s3 θ - Cos θ
= Sinθ( 1 - 2Sin2 θ)
Cos θ (2C0s2 θ - 1 )
= Sinθ [ 1 – 2( 1- Cos2θ)]
Cos θ (2C0s2 θ - 1 )
= Sinθ( 1 – 2 + 2Cos2θ)
Cos θ (2C0s2 θ - 1 )
= Sinθ ( 2Cos2θ - 1 )
Cos θ (2C0s2 θ - 1 )
= Sinθ
Cosθ
= Tanθ
= RHS
Hence it is proved.

17. Prove that :
Cosθ - 2Cos3 θ = Cotθ
2Sin3 θ - Sin θ

18. VIII)
(SinA+CosecA) 2 + (CosA + Sec A) 2 = 7 + Tan2A +Cot2A
Solution:
(SinA+CosecA) 2 = Sin 2 A+Cosec2 A + 2 . SinA.CosecA
= Sin 2A+ 1+ Cot2 A + 2 . SinA. 1/Sin A
= Sin 2A+ Cot2 A + 3 ------1
(CosA + Sec A) 2 = Cos2 A + Sec2 A + 2 . CosA.SecA
= Cos2 A +1 + Tan2A + 2.CosA.1/Cos A
= Cos2 A + Tan2A + 3 ------ 2
now, LHS =1+2 = Sin 2A+ Cot2 A+3 + Cos2 A+Tan2A + 3
= 1 + 3 + 3 + Tan2A +Cot2A
= 7 + Tan2A +Cot2A = RHS
Hence it is proved.

19. x)(1 + Tan2A)/(1+ Cot2 A)= [(1 - TanA)/(1-CotA)]2 = Tan2A
Solution:
(1 + Tan2A) / (1+ Cot2 A) = Sec2A / Cosec2A
= (1/Cos2A) / ( 1/Sin2A)
= 1/Cos2A x Sin2A/1
= Sin2A /Cos2A = Tan2A
[(1 + TanA) / (1+CotA)]2 = [(1 - SinA/CosA) / (1- CosA/SinA)]2
= [(CosA- SinA)/CosA /(SinA -
CosA)/SinA)]2
= [(CosA+ SinA)/CosA X - (SinA
/(CosA +SinA)]
= ( - SinA / Cos A ) 2 = (-
TanA ) 2 = Tan 2 A therefore,
(1 + Tan2A)/(1+ Cot2 A)= [(1 - TanA)/(1-CotA)]2 = Tan2A
Hence it is proved.

20. Prove that:
(1 + Cot2A) / (1+ Tan2 A)
= [(1 - Cot A) / (1- TanA)]2
= Cot2A

21. Prove that √( 1 + Sin A ) / √( 1 – SinA ) = Sec A + TanA
LHS = √( 1 + Sin A ) / √( 1 – SinA )
Rationalising the denominator
= √( 1 + Sin A ) / √( 1 – SinA ) x √( 1 + Sin A ) / √( 1 + SinA )
= √( 1 + Sin A ) √( 1 + Sin A ) / √( 1 – SinA ) √( 1 + Sin A )
= √( 1 + Sin A ) 2 / √( 1 - Sin2 A )
= (1 + Sin A ) / √ Cos2 A
= (1 + Sin A ) / Cos A
= 1 / CosA + Sin A/ Cos A
= Sec A + Tan A = RHS Hence it proved.

22. Prove that:
√( 1 + Cos A ) / √( 1 – CosA ) = Cosec A + CotA