Question

# Prove that: (i)tan225∘cot405∘+tan765∘cot675∘=0 (ii)sin8π3cos23π6+cos13π3sin35π6=12 (iii)cos24∘+cos55∘+cos125∘+cos204∘+cos300∘=12 (iv)tan(−225∘)cot(−405∘)−tan(−765∘)cot(675∘)=0 (v)cos570∘sin510∘+sin(−330∘)cos(−390∘)=0 (vi)tan11π3−2sin4π6−34cosec2π4+4cos217π6=3−4√32 (vii)3sinπ6secπ3−4sin5π5cotπ4=1

Solution

## (i)LHS=tan225∘cot405∘+tan765∘cot675∘ =tan(π+π4)cot(2π+π4)+tan(4π+π4)cot(4π−π4) =tanπ4.cotπ4+tanπ4×(−cotπ4) [∵cot(4π−π4)=−cotπ4] =1.1 +1.(-1) =RHS Hence proved. (ii)LHS=sin8π3cos23π6+cos13π3sin35π6 =sin(3π−π3)cos(4π−π6)+cos(4π+π3)sin(6π−π6) =sinπ3cosπ6+cosπ3(−sinπ6) [∵sin(6π−θ)=−sinθ] =√32×√322+12×(−12) =34−14 =24 =12=RHS Hence proved. (iii)LHS=cos24∘+cos55∘+cos125∘+cos204∘+cos300∘ =cos24∘+cos204∘+cos55∘+cos125∘+cos300∘ =cos24∘+cos(π+24∘)+cos55∘+cos(π−55∘)+cos(2π−π3) =cos24∘−cos24∘+cos55∘−cos55∘+cosπ3 =cosπ3 =12=RHS Hence proved. (iv)LHS=tan(−225∘)cot(−405∘)−tan(−765∘)cot(675∘) =−tan225∘(−cot405∘)+tan765∘cot765∘ [∵tan(−θ)=−tanθ and cot(−θ)=−cotθ] =tan(π+π4)cot(2ππ4)+tan(4π+π4)cot(4π−π4) =tanπ4cotπ4+tanπ4×(−cotπ4)[∵cot(4π−θ)=−cotθ] =1.1+1(-1)=1-1=0=RHS Hence proved. (v)LHS=cos 570∘ sin 510∘+sin(−330∘)cos(−390∘) =cos(3π+π6)sin(3π−π6)−sin330∘cos390∘ [∵sin(−θ)=−sinθandcos(−θ)=cosθ] =−cosπ6sinπ6−sin(2π−π6)cos(2π+π6) =−sinπ6cosπ6+sinπ6cosπ6[∵sin(2π−θ)=−sinθ] 0=RHS Hence proved. (vi)LHS=tan11π3−2sin4π6−34cosec2π4+4cos217π6 =tan(4π−π3)−2sin2π3−34×(√2)2+4cos2(3π−π6) =−tanπ3−2sin(π−π3)−34×2+4cos2π6 (∵tan(4π−π3)=−tanπ3,cos(3π−π6)=−cosπ6) =−√3−2sinπ3−32+4×(√32)2 =−√3−2×√32−32+4×34 =−√3−√3−32+3=−2√3−3+62=−2√332 =3−4√32 =RHS Hence proved. (vii)LHS=3sinπ6secπ3−4sin5π6cotπ4 =3×12×2−4sin(π−π6)×1 =3−4siinπ6[∵sin(π−θ)=sinθ] =3−4×12=3−2=1 =RHS Hence proved. MathematicsRD SharmaStandard XI

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