Class 10 Mathematics

# Trigonometry

## Exercise 8.4 (NCERT) Part 2

Question – 5 - Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(i) (text(cosec)θ-cotθ)^2=(1-cosθ)/(1+cosθ)

Solution:

LHS: (text(cosec)θ-cotθ)^2

=((1)/(sinθ)-(cosθ)/(sinθ))^2

=((1-cosθ)/(sinθ))^2

=((1-cosθ)^2)/(sin^2θ)

RHS: (1-cosθ)/(1+cosθ)

=(1-cosθ)/(1+cosθ)xx(1-cosθ)/(1-cosθ)

((1-cosθ)^2)/(1-cos^2θ)

=((1-cosθ)^2)/(sin^2θ)

LHS=RHS proved

(ii) (cosA)/(1+sinA)+(1+sinA)/(cosA)=2secA

Solution:

LHS: (cosA)/(1+sinA)+(1+sinA)/(cosA)

=(cos^2A+sin^2A+2sinA+1)/((1+sinA)cosA)

=(1+2sinA+1)/((1+sinA)cosA)

=(2+2sinA)/((1+sinA)cosA)

=(2(1+sinA))/(cosA(1+sinA))

=(2)/(cosA)=2secA

(iii) (tanθ)/(1-cotθ)+(cotθ)/(1-tanθ) =1+secθtext(cosec)θ

Solution:

LHS: (tanθ)/(1-cotθ)+(cotθ)/(1-tanθ)

=(tanθ)/(1-(cosθ)/(sinθ))+(cotθ)/(1-(sinθ)/(cosθ))

=((sinθ)/(cosθ))/((sinθ-cosθ)/(sinθ))+((cosθ)/(sinθ))/((cosθ-sinθ)/(cosθ))

=(sinθ)/(cosθ)xx(sinθ)/(sinθ-cosθ)+(cosθ)/(sinθ)xx(cosθ)/(cosθ-sinθ)

=(sin^2θ)/(cosθ(sinθ-cosθ))+(cos^2θ)/(sinθ(cosθ-sinθ))

=(sin^2θ)/(cosθ(sinθ-cosθ))+(cos^2θ)/(sinθ(-sinθ+cosθ))

=(sin^2θ)/(cosθ(sinθ-cosθ))-(cos^2θ)/(sinθ(sinθ-cosθ))

=(1)/(sinθ-cosθ)((sin^2θ)/(cosθ)-(cos^2θ)/(sinθ))

=(1)/(sinθ-cosθ)((sin^3θ-cos^3θ)/(sinθ.cosθ))

=(1)/(sinθ-cosθ)xx((sinθ-cosθ)(sin^2θ+cos^2θ+sinθ.cosθ))/( sinθ.cosθ)

=(sin^2θ+cos^2θ+sinθ.cosθ)/(sinθ.cosθ)

=(1+ sinθ.cosθ)/( sinθ.cosθ)

=(1)/( sinθ.cosθ)+( sinθ.cosθ)/( sinθ.cosθ)

=(1)/( sinθ.cosθ)+1

=secθ.text(cosec)θ+1

=1+sec&theta.text(cosec)θ=RHS proved

(iv) (1+secA)/(secA)=(sin^2A)/(1-cosA)

Solution:

LHS: (1+secA)/(secA)

=(1+(1)/(cosA))/((1)/(cosA))

=((cosA+1)/(cosA))/((1)/(cosA))

=1+cosA

=1+cosAxx(1-cosA)/(1-cosA)

=(1-cos^2A)/(1-cosA)

=(sin^2A)/(1-cosA)=RHS

(v) (cosA-sinA+1)/(cosA+sinA-1)=text(cosec)A+cotA

using the identity cosec2 A = 1 + cot2 A

Solution:

LHS: (cosA-sinA+1)/(cosA+sinA-1)

After dividing numerator and denominator by sin A, we get

((cosA-sinA+1)/(sinA))/((cosA+sinA-1)/(sinA))

=((cosA)/(sinA)-(sinA)/(sinA)+(1)/(sinA))/ ((cosA)/(sinA)+(sinA)/(sinA)-(1)/(sinA))

=(cotA-1+text(cosec)A)/(cotA+1-text(cosec)A)

After multiplying with [cotA – (1-cosecA)] in numerator and denominator both we get

=(cotA-1+text(cosec)A)/(cotA+1-text(cosec)A)xx((cotA)-(1-text(cosec)A))/((cotA)-(1-text(cosec)A)

=((cotA)-(1-text(cosec)A))/((cotA)+(1-text(cosec)A))xx(((cotA)-(1-text(cosec)A))/( ((cotA)-(1-text(cosec)A))

=([(cotA)-(1-text(cosec)A)]^2)/((cotA)^2-(1-text(cosec)A)^2)

=(cot^2+(1-text(cosec)A)^2-2xx\cotA(1-text(cosec)A))/(cot^2A-(1+text(cosec)^2A-2text(cosec)A))

=(cot^2A+1+text(cosec)^2A-2text(cosec)A-2cotA+2cotA.text(cosec)A)/(cot^2A-1-text(cosec)^2A+2text(cosec)A)

=((cot^2A+1)+text(cosec)^2A-2text(cosec)A-2cotA+2cotA.text(cosec)A)/(-(cot^2A-+text(cosec)^2A)-1+2text(cosec)A)

After using the identity; cosec2 A = 1 + cot2A, we get;

(text(cosec)^2A+text(cosec)^2A-2text(cosec)A-2cotA+2cotA.text(cosec)A)/(-1-1+2text(cosec)A)

=(2text(cosec)^2A+2cotA.text(cosec)A-2cotA-2text(cosec)A)/(-2+2text(cosec)A)

=(2text(cosec)A(text(cosec)A+cotA)-2(cotA+text(cosec)A))/(2text(cosecA-2)

=((2text(cosec)A-2)(text(cosec)A+cotA))/(2text(cosec)A-2)

=text(cosec)A+cotA=RHS

(vi) sqrt((1+sinA)/(1-sinA))=secA+tanA

Solution:

LHS: sqrt((1+sinA)/(1-sinA))

=sqrt((1+sinA)/(1-sinA)xx(1+sinA)/(1+sinA))

=sqrt(((1+sinA)^2)/(1-sin^2A))

=sqrt(((1+sinA)^2)/(cos^2A))

=(1+sinA)/(cosA)

=(1)/(cosA)+(sinA)/(cosA)

=secA+tanA=RHS

(vii) (sinθ-2sin^3θ)/(2cos^3θ-cosθ)=tanθ

Solution:

LHS: (sinθ-2sin^3θ)/(2cos^3θ-cosθ)

=(sinθ(1-2sin^2θ))/(cosθ(2cos^2θ-1))

=(sinθ[1-2(1-cos^2&theta)])/(cosθ(2cos^2θ-1))

=(sinθ(1-2+2cos^2θ))/(cosθ(2cos^2θ-1))

=(sinθ(2cos^2θ-1))/(cosθ(2cos^2θ-1))

=(sinθ)/(cosθ)=tanθ=RHS

(viii) (sinA+text(cosec)A)^2+(cosA+secA)^2=7+tan^2A+cot^2A

Solution:

LHS: (sinA+text(cosec)A)^2+(cosA+secA)^2

=sin^2A+text(cosec)^2A+2sinA.text(cosec)A+cos^2A+sec^2A+2cosA.secA

=sin^2A+cos^2A+text(cosec)^2A+sec^2A+2+2

=1+2+2+text(cosec)^2A+sec^2A

=5+(1)/(sin^2A)+(1)/(cos^2A)

=5+(cos^2A+sin^2A)/(sin^2A.cos^2A)

=5+(1)/(sin^2A.cos^2A)

Now;

RHS: 7+tan^2A+cot^2A

=7+(sin^2A)/(cos^2A)+(cos^2A)/(sin^2A)

=7+(sin^4A+cos^4A)/(sin^2A.cos^2A)

=7+((sin2A+cos^2A)^2-2sin^2A.cos^2A)/(sin^2A.cos^2A)

=7+(1+2sin^2A.cos^2A)/(sin^2A.cos^2A)

=(7 sin^2A.cos^2A+1-2 sin^2A.cos^2A)/( sin^2A.cos^2A)

=(5 sin^2A.cos^2A+1)/( sin^2A.cos^2A)

=5+(1)/( sin^2A.cos^2A)=LHS

Hence; LHS = RHS proved

(ix) (text(cosec)A-sinA)(secA-cosA)=(1)/(tanA+cotA)

Solution:

LHS: (text(cosec)A-sinA)(secA-cosA)

=((1)/(sinA)-sinA)((1)/(cosA)-cosA)

=((1-sin^2A)/(sinA))((1-cos^2A)/(cosA))

=(cos^2A)/(sinA)xx(sin^2A)/(cosA)

=sinA.cosA

RHS: (1)/(tanA+cotA)

=(1)/((sinA)/(cosA)+(cosA)/(sinA))

=(1)/((sin^2A+cos^2A)/(sinA.cosA))

=sinA.cosA=LHS

(x) (1+tan^2A)/(1+cot^2A)=((1-tanA)/(1-cotA))^2=tan^2A

Solution:

LHS: (1+tan^2A)/(1+cot^2A)

=(1+tan^2A)/(1+(1)/(tan^2A))

=(1+tan^2A)/((1+tan^2A)/(tan^2A))

=tan^2A

Middle Term: ((1-tanA)/(1-cotA))^2

=((1-tanA)/(1-(1)/(tanA)))^2

=((1-tanA)/((tanA-1)/(tanA)))^2

=(1+tan^2A-2tanA)/((1+tan^2A-2tanA)/(tan^2A))

=tan^2A

Hence; LHS = Middle Term = RHS proved

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Exercise 8.1(part I)

Exercise 8.1(part II)

Exercise 8.2

Exercise 8.3

Exercise 8.4(part I)

Application(part I)

Application(part II)