If cosxcos(x-2y)=λ, then tan(x-y)tany is equal to
(1+λ)(1–λ)
(1-λ)(1+λ)
λ(1+λ)
λ(1–λ)
Explanation for the correct option:
Step 1. Find the value of tan(x-y)tany:
Given, cosxcos(x-2y)=λ
Now,
tan(x–y)tany=sin(x–y)cos(x–y)sinycosy=2sin(x–y)siny2cos(x–y)cosy
Step 2. Use the formulas 2sinCsinD=cos(C–D)–cos(C+D)and 2cosCcosD=cos(C+D)+cos(C–D)
=cos(x–y–y)–cos(x–y+y)cos(x–y+y)+cos(x–y–y)=cos(x–2y)–cosxcosx+cos(x–2y)
Step 3. Take and cancel cos(x–2y) from numerator and denominator, we get
=1–cosxcos(x–2y)cosxcos(x–2y)+1=(1–λ)(λ+1)=(1–λ)(1+λ)
Hence, Option ‘B’ is Correct.