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Question

Prove that:

(i) sinθ cos(90°-θ)+sin(90°-θ) cosθ=1
(ii) sinθcos(90°-θ)+cosθsin(90°-θ)=2
(iii) sinθ cos(90°-θ)cosθsin(90°-θ)+cosθ sin(90°-θ)sinθcos(90°-θ)=1
(iv) cos(90°-θ)sec(90°-θ)tanθcosec(90°-θ)sin(90°-θ)cot(90°-θ)+tan(90°-θ)cotθ=2
(v) cos(90°-θ)1+sin(90°-θ)+1+sin(90°-θ)cos(90°-θ)=2cosecθ
(vi) sec90°-θ cosecθ-tan90°-θ cotθ+cos225°+cos265°3tan27° tan63°=23 CBSE 2010
(vii) cotθ tan90°-θ-sec90°-θcosecθ+3tan12° tan60° tan78°=2 CBSE 2010

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Solution

(i) LHS=sinθcos(900θ)+sin(900θ)cosθ =sinθsinθ+cosθcosθ =sin2θ+cos2θ =1 = RHS Hence proved.(ii) LHS=sinθcos(900θ)+cosθsin(900θ) =sinθsinθ+cosθcosθ =1+1 =2 =RHS Hence proved. (iii) LHS=sinθcos(900θ)cosθsin(900θ)+cosθsin(900θ)sinθcos(900θ) =sinθsinθcosθcosθ+cosθcosθsinθsinθ =sin2θ+cos2θ =1=RHS Hence proved(iv) LHS=cos(900θ)sec(900θ)tanθcosec(900θ)sin(900θ)cot(900θ)+tan(900θ)cotθ =sinθcosecθtanθsecθcosθtanθ+cotθcotθ =1+1 =2 =RHS Hence proved.(v) LHS=cos(900θ)1+sin(900θ)+1+sin(900θ)cos(900θ) =sinθ1+cosθ+1+cosθsinθ =sin2θ+(1+cosθ)2(1+cosθ)sinθ =sin2θ+1+cos2θ+2cosθ(1+cosθ)sinθ =1+1+2cosθ(1+cosθ)sinθ =2+2cosθ(1+cosθ)sinθ =2(1+cosθ)(1+cosθ)sinθ =21sinθ =2cosecθ = RHS Hence proved.

vi LHS=sec90°-θ cosecθ-tan90°-θ cotθ+cos225°+cos265°3tan27° tan63°=cosecθ cosecθ-cotθ cotθ+sin290°-25°+cos265°3tan27° cot90°-63°=cosec2θ-cot2θ+sin265°+cos265°3tan27° cot27°=1+13×tan27°×1tan27°=23=RHS

vii LHS=cotθ tan90°-θ-sec90°-θcosecθ+3tan12° tan60° tan78°=cotθ cotθ-cosecθ cosecθ+3tan12°×3×cot90°-78°=cot2θ-cosec2θ+3tan12° cot12°=-1+3×tan12°×1tan12°=-1+3=2=RHS

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