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Question

Prove that:
(i) sinA+sin3AcosAcos3A=cotA
(ii) sin9Asin7Acos7Acos9A=cot8A
(iii) sinAsinBcosAcosB=tanAB2
(iv) sinA+sinBsinAsinBtan(A+B2)cotA+B2
(v) cosA+cosBcosBcosA=cotA+B2cotAB2

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Solution

(i) sinA+sin3AcosAcos3A=cotALHS=sinA+sin3AcosAcos3A=cotA=2sin(A+3A2)cot(A3A2)sin(A+3A2)sin(A3A2)=sin2A×cos(A)sin2Asin(A)=cos(A)sin(A)=cosAsinA=cosAsinA=cotA=RHSsinA+sin3AcosAcos3A=cotA

(ii) sin9Asin7Acos7Acos9A=cot8A

We have,
LHS=sin9Asin7Acos7sAcos9A=2sin(9A7A2)cos(9A+7A2)2sin(7A+9A2)sin(7A9A2)=sinA cos8Asin8A×sin8A [sin(θ)=sinθ]=cos8Asin8A=cot8A=RHS sin9A+sin7Acos7Acos9A

(iii) sinAsinBcosAcosB=tanAB2
We have,
LHS=sinAsinBcos+cosB=2cos(A+B2)sin(AB2)2cos(A+B2)cos(AB2)=sin(AB2)cos(AB2)=RHS sinAsinBcosA+cosB=tan(AB2)

We have,
(iv) LHS=sinA+sinBsinAsinB=2sin(A+B2)cos(AB2)2sin(AB2)cos(A+B2)=sin(A+B2)cos(AB2)sin(A+B2)sin(AB2)=tan(A+B2)cotsin(AB2)=RHS sinA+sinBsinAsinB=tan(A+B2)cot(AB2)

(v) We have,
LHS=cosA+cosBcosBcosA=2cos(A+B2)cos(AB2)2sin(B+A2)sin(BA2)=cos(A+B2)cos(AB2)sin(A+B2)sin(BA2) [sin(θ)=sinθ]=cos(A+B2)cos(AB2)sin(A+B2)sin(AB2)=cot(A+B2)cot(AB2)=RHS cosA+cosBcosBcosA=cot(A+B2)cot(AB2)


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