"The mathematician is fascinated with the marvelous beauty of the forms he constructs, and in their beauty he finds everlasting truth." If xcosθ – ysinθ = a, xsinθ + ycos θ = b, prove that x2+y2=a2+b2.

1.

Ans: xcosθ - y sinθ = a xsinθ + y cosθ = b Squaring and adding x2+y2=a2+b2. 2. Prove that sec2θ+cosec2θ can never be less than 2.

Ans: S.T Sec2θ + Cosec2θ can never be less than 2. If possible let it be less than 2. 1 + Tan2θ + 1 + Cot2θ < 2. ⇒ 2 + Tan2θ + Cot2θ ⇒ (Tanθ + Cotθ)2 < 2. Which is not possible. 3. If sinϕ = , show that 3cosϕ-4cos3ϕ = 0.

Ans: Sin ϕ = ½ ⇒ ϕ = 30o Substituting in place of ϕ =30o. We get 0. 4. If 7sin2ϕ+3cos2ϕ = 4, show that tanϕ = 1 .

Ans: If 7 Sin2ϕ + 3 Cos2ϕ = 4 S.T. Tanϕ

2 2 2

3 7 Sin ϕ + 3 Cos ϕ = 4 (Sin ϕ + Cos ϕ)

2

⇒ 3 Sin2ϕ = Cos2ϕ ⇒ Sin 2ϕ 1 = Cos 2ϕ 3

43

⇒ Tan2ϕ =

1 3

1 3

Tanϕ =

5.

If cosϕ+sinϕ =

cosϕ, prove that cosϕ - sinϕ =

sin ϕ.

Ans: Cosϕ + Sinϕ = 2 Cosϕ ⇒ ( Cosϕ + Sinϕ)2 = 2Cos2ϕ ⇒ Cos2ϕ + Sin2ϕ+2Cosϕ Sinϕ = 2Cos2ϕ ⇒ Cos2ϕ - 2Cosϕ Sinϕ+ Sin2ϕ = 2Sin2ϕ ⇒ (Cosϕ - Sinϕ)2 = 2Sin2ϕ Cos2ϕ or Cosϕ - Sinϕ = 2 Sinϕ. 6.

∴2Sin2ϕ = 2 - 2Cos2ϕ 1- Cos2ϕ = Sin2ϕ & 1 - Sin2ϕ =

If tanA+sinA=m and tanA-sinA=n, show that m2-n2 = 4

Ans: TanA + SinA = m TanA – SinA = n. 2 2 m -n =4 mn . m2-n2= (TanA + SinA)2-(TanA - SinA)2 = 4 TanA SinA RHS 4 mn = 4 (TanA + SinA)(TanA − SinA) =4 =4 =4

Tan 2 A − Sin 2 A

Sin 2 A − Sin 2 ACos 2 A Cos 2 A

Sin 4 A Cos 2 A Sin 2 A = 4 TanA SinA =4 Cos 2 A ∴m2 – n2 = 4 mn

7.

If secA=

, prove that secA+tanA=2x or

.

44

Ans: Secϕ = x +

1 4x (Sec2ϕ= 1 + Tan2ϕ)

⇒ Sec2ϕ =( x +

1 2 ) 4x 1 2 Tan2ϕ = ( x + ) -1 4x 1 2 ) Tan2ϕ = ( x 4x Tanϕ = + x 1 4x

Secϕ + Tanϕ = x + = 2x or 8. 1 2x

1 1 + x4x 4x

If A, B are acute angles and sinA= cosB, then find the value of A+B.

Ans: A + B = 90o

a)Solve for ϕ, if tan5ϕ = 1.

9.

Ans: Tan 5ϕ = 1

⇒ϕ=

45 ⇒ ϕ=9o. 5

b)Solve for ϕ if

Sinϕ 1 + Cosϕ + = 4. 1 + Cosϕ Sinϕ

Ans:

Sinϕ 1 + Cosϕ + =4 1 + Cosϕ Sinϕ

Sin 2 ϕ + 1(Cosφ) 2 =4 Sinϕ(1 + Cosϕ)

Sin 2 ϕ + 1 + Cos 2 ϕ + 2Cosϕ =4 Sinϕ + SinϕCosϕ

2 + 2Cosϕ =4 Sinϕ (1 + Cosϕ )

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⇒

2 + (1 + Cosϕ ) =4 Sinϕ (1 + Cosϕ ) 2 =4 Sinϕ Sinϕ = 1 2 Sinϕ = Sin30 ϕ = 30o

⇒

⇒ ⇒

10.

If

Ans:

Cosα =m Cosβ Cos 2α ⇒m Cos 2 β

2=

Cosα =n Sinβ

n

2= Cos

α Sin β

2 2

LHS = (m2+n2) Cos2 β

Cos 2α Cos 2α 2 + Cos β Cos 2 β Sin 2 β 1 Cos 2α Cos 2 βSin 2 β = Cos 2α 2 =n Sin 2 β Cos 2 β

=

⇒(m2+n2) Cos 2 β =n2

11.

If 7 cosecϕ-3cotϕ = 7, prove that 7cotϕ - 3cosecϕ = 3.

Ans: 7 Cosecϕ-2Cotϕ=7 P.T 7Cotϕ - 3 Cosecϕ=3 7 Cosecϕ-3Cotϕ=7 ⇒7Cosecϕ-7=3Cotϕ ⇒7(Cosecϕ-1)=3Cotϕ

46

⇒7(Cosecϕ-1) (Cosecϕ+1)=3Cotϕ(Cosecϕ+1) ⇒7(Cosec2ϕ-1)=3Cotϕ(Cosecϕ+1) ⇒7Cot2ϕ=3 Cotϕ (Cosecϕ+1) ⇒7Cotϕ= 3(Cosecϕ+1) 7Cotϕ-3 Cosecϕ=3 12. 2(sin 6ϕ+cos6ϕ) – 3(sin4ϕ+cos4ϕ)+1 = 0

Ans: (Sin2ϕ)3 + (Cos2ϕ)3-3 (Sin4ϕ+(Cos4ϕ)+1=0 Consider (Sin2ϕ)3 +(Cos2ϕ)3 ⇒(Sin2ϕ+Cos2ϕ)3-3 Sin2ϕCos2ϕ( Sin2ϕ+Cos2ϕ) = 1- 3Sin2ϕ Cos2ϕ Sin4ϕ+Cos4ϕ(Sin2ϕ)2+(Cos2ϕ)2 = (Sin2ϕ+Cos2ϕ)2-2 Sin2ϕ Cos2ϕ = 1- 2 Sin2ϕ Cos2ϕ = 2(Sin6ϕ+Cos6ϕ)-3(Sin4ϕ+Cos4ϕ) +1 = 2 (1-3 Sin2ϕ Cos2ϕ)-3 (1-2 Sin2ϕ+Cos2ϕ)+1 13. 5(sin8A- cos8A) = (2sin2A – 1) (1- 2sin2A cos2 A)

Ans: Proceed as in Question No.12

5 & θ +φ =90o what is the value of cotφ. 6 5 Ans: Tanθ = i.e. Cotφ = 5 Since ϕ + θ = 90o. 6 6 14. If tanθ = What is the value of tanϕ in terms of sinϕ. Sinϕ Ans: Tan ϕ = Cosϕ Sinϕ Tan ϕ = 1 − Sin 2ϕ 16. If Secϕ+Tanϕ=4 find sin ϕ, cosϕ 15.

Ans: Sec ϕ + Tan ϕ = 4

1 Sinϕ + =4 Cosϕ Cosϕ 1 + Sinϕ =4 Cosϕ

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(1 + Sinϕ ) 2 ⇒ = 16 Cos 2ϕ

⇒ apply (C & D)

= (1 + Sinφ ) 2 + Cos 2φ 16 + 1 = (1 + Sinφ ) 2 − Cos 2φ 16 − 1

2(1 + Sinφ ) 17 = 2 Sinφ (1 + Sinφ ) 15 1 17 = ⇒ Sinφ 15 15 ⇒Sinϕ= 17

⇒

Cosϕ = 1 − Sin 2ϕ

8 15 1− = 17 17 17. Secϕ+Tanϕ=p, prove that sinϕ = p2 −1 p2 +1

2

P2 −1 Ans: Secϕ + Tanϕ= P. P.T Sinϕ= 2 P +1 Proceed as in Question No.15

18....