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)