Question

# Prove that: (i)tan720∘−cos270∘−sin150∘cos120∘=14 (ii)sin780∘sin480∘+cos120∘sin150∘=12 (iii)sin780∘sin120∘+cos240∘sin390∘=12 (iv)sin600∘cos390∘+cos480∘sin150∘=−1 (v)tan250∘cot405∘+tan765∘cot675∘=0

Solution

## (i)LHS=tan720∘−cos270∘−sin150∘cos120∘ =tan4π−cos(3π2)−sin(π−π6) cos(π2+π6)[∵π=180∘] =0−0−sinπ6(−sinπ6) [∵tannπ=0foralln∈Zandcos3π2=0] =sin2π6 =(12)2=14=RHS Hence proved. (ii)LHS=sin780∘sin480∘+cos120∘sin150∘ =sin(4π+π3)sin(3π−π3)+cos (π2+π6)sin(π−π6) [∵π=180∘] =sinπ3×sinπ3+(−sinπ6)sinπ6 ⎡⎢⎣∵sin(4π+π3)=sinπ3and sin(3π+π3)=sinπ3⎤⎥⎦ =√32×√32−12×12=34−14 =24=12=RHS Hence proved. (iii)LHS=sin780∘sin120∘+cos240∘sin390∘ =sin(4π+π3)sin(π2+π6)+cos(π+π6)sin(2π+π6) =sinπ3×cosπ6−cosπ3×(+sinπ6) =√32×√32−12×12 =34−14 =12=12=RHS Hence proved. (iv)LHS=sin600∘cos390∘+cos480∘sin150∘ =sin(3π+π3)cos(2π+π6)+cos(3π−π3)sin(π−π6) =−sinπ3cosπ6−cosπ3−sinπ6 ⎡⎢⎣∵sin(3π+π3)=−sinπ3 and cos(3π−π3)=−cosπ3⎤⎥⎦ =−√32×−sqrt32−12×12=−34−14=−44 =-1 =RHS Hence proved.= (v)LHS=tan225∘cot405∘+tan765∘cot675∘ =tan(π+π4)cot(2π+π4)(4π+π4)cot(4π−π4) =tanπ4cotπ4+tanπ4(−cotπ4) =1.1 +1.(-1)=0 RHS Hence proved.MathematicsRD SharmaStandard XI

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