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# Prove that: (i) sinA+sin3AcosA−cos3A=cotA (ii) sin9A−sin7Acos7A−cos9A=cot8A (iii) sinA−sinBcosA−cosB=tanA−B2 (iv) sinA+sinBsinA−sinBtan(A+B2)cotA+B2 (v) cosA+cosBcosB−cosA=cotA+B2cotA−B2

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Solution

## (i) sinA+sin3AcosA−cos3A=cotALHS=sinA+sin3AcosA−cos3A=cotA=2sin(A+3A2)cot(A−3A2)−sin(A+3A2)sin(A−3A2)=−sin2A×cos(−A)sin2Asin(−A)=−cos(−A)sin(−A)=−cosA−sinA=cosAsinA=cotA=RHS∴sinA+sin3AcosA−cos3A=cotA (ii) sin9A−sin7Acos7A−cos9A=cot8A We have, LHS=sin9A−sin7Acos7sA−cos9A=2sin(9A−7A2)cos(9A+7A2)−2sin(7A+9A2)sin(7A−9A2)=−sinA cos8Asin8A×sin8A [∵sin(−θ)=−sinθ]=cos8Asin8A=cot8A=RHS∴ sin9A+sin7Acos7A−cos9A (iii) sinA−sinBcosA−cosB=tanA−B2 We have, LHS=sinA−sinBcos+cosB=2cos(A+B2)sin(A−B2)2cos(A+B2)cos(A−B2)=sin(A−B2)cos(A−B2)=RHS∴ sinA−sinBcosA+cosB=tan(A−B2) We have, (iv) LHS=sinA+sinBsinA−sinB=2sin(A+B2)cos(A−B2)2sin(A−B2)cos(A+B2)=sin(A+B2)cos(A−B2)sin(A+B2)sin(A−B2)=tan(A+B2)cotsin(A−B2)=RHS∴ sinA+sinBsinA−sinB=tan(A+B2)cot(A−B2) (v) We have, LHS=cosA+cosBcosB−cosA=2cos(A+B2)cos(A−B2)2sin(B+A2)sin(B−A2)=cos(A+B2)cos(A−B2)sin(A+B2)sin(B−A2) [∵sin(−θ)=−sinθ]=−cos(A+B2)cos(A−B2)−sin(A+B2)sin(A−B2)=cot(A+B2)cot(A−B2)=RHS∴ cosA+cosBcosB−cosA=cot(A+B2)cot(A−B2)

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