The number of values of θ satisfying 4cosθ+3sinθ=5 as well as 3cosθ+4sinθ=5 is
A
1
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B
2
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C
0
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D
None of these
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Solution
The correct option is C0 4cosθ+3sinθ=5 and 3cosθ+4sinθ=5 ⇒45cosθ+35sinθ=1 Let sinA=45∴cosA=35 ∴sinAcosθ+cosAsinθ=1 ⇒sin(A+θ)=1 ⇒A+θ=π2.....(1) Now second equation 3cosθ+4sinθ=5 35cosθ+45sinθ=1 ⇒cosAcosθ+sinAsinθ=1 ⇒cos(θ−A)=1 ⇒θ−A=0 ⇒θ=A ∴2θ=π2 ........ From (1) ⇒θ=π4 Put value of θ in 4cosθ+3sinθ=5 LHS= 4cosπ4+3sinπ4=4√2+3√2=7√2≠5 ∴ No solution exists No. of solutions =0